The short answer is “no.” The long answer is, “it depends on what you intend to do with the ranking.”
First, let us ignore asset classes because you are assuming something that is a priori impossible. We never know the actual population parameters. If we did, we would not need the field of statistics. Nonetheless, we can discuss this question as independent of economics or finance. We can approach this as a statistics question.
You will need to define some things, such as the probability distributions involved and the school of thought, before proceeding. Not all probability distributions have a covariance matrix. In those cases, a correlation may imply something other than what you intend it to mean. You will also need to define a loss function. There may also be other things that may have to be declared, such as the relationships among samples.
Let us begin with a simple case where we have $n$ sets of samples. None of these samples contain overlapping data. Each sample is independent. The parameters are stationary. The set contains variables $X_1\dots{X_k}$. The parameter set contains population parameters $\rho_{i,j}$ where $i$ and $j$ denote separate assets, and $\rho$ is the population parameter of the correlation between $X_i$ and $X_j$. For simplicity, let us also assume that $\rho_{i,j}=\rho_{j,i}$. For some ideas of association, such as those measured by Somers’ D, $\rho_{i,j}\ne\rho_{j,i}$.
We need some form of a loss function. In a decision-theoretic construction, you should choose your own utility function such that $$-U(\rho,\hat{\rho})=L(\rho,\hat{\rho}),$$ where $U$ is your utility function and $L$ is your loss function. You can also adopt a loss function that serves some other theoretical purpose, such as quadratic loss. For our purpose, we could assume quadratic loss. That assumption may not always work, depending on the distributions involved. A careful study should precede your choices in cases where that will not work, such as with many distributions with heavy tails.
Our loss will be defined as $\sum_{\forall{i\ne{j}}}c(\rho_{i,j}-\hat{\rho_{i,j}})^2,c>0.$
For Frequentist methods, Pearson’s rho is the loss minimizing estimator. For Bayesian methods, the posterior mean of the correlation coefficient will minimize the loss. For Likelihoodist methods, the solution is the method of maximum likelihood for most distribution and problems. Implicitly, we have to assume that the probability distributions involved have a finite second moment. In assuming quadratic loss, we have excluded a wide class of real-world probability distributions. One could assume an alternative loss function if that were to happen; however, it would also remove both Pearson’s $\rho$ and the posterior mean as a loss minimizing tool. Some other set of metrics would minimize those alternative loss functions.
Now let us get to your question.
Automatically, if we create ranks $Y_1\dots{Y_n}$, where $Y_q=rank(\sum_{\forall{i\ne{j}}}c(\rho_{i,j}-\hat{\rho_{i,j}})^2),q\in{1\dots{n}}$, then you have your solution.
The problem is that it does not tell you anything with meaning to it. To consider why, imagine the following problem.
You are measuring the height of fifth-grade boys to estimate the population mean. You have randomly selected the Bruno de Finetti School for Boys to make your estimate at. We will eliminate any potential methodological issues away through the use of our magic wand.
Despite our magical powers, a methodological dispute has broken out and has been solved by randomly splitting the students into two groups of equal size using a perfectly fair coin. It is fortunate that we have an even number of students so that no chainsaw was required.
The first group’s sample average was 55 inches tall, and the standard deviation was 2 inches. The second group had a sample average of 55.1 inches and a standard deviation of 1.9 inches. Could we say that our second group had a more precise estimate since the confidence interval was narrower?
No, we definitely could not do so because the treatment was the random toss of a coin, and nothing about random selection can improve precision. Even though you could rank the two groups in terms of narrowness, there is no meaning at all in the ranking due to the very definition of a confidence interval.
The same thing is true for a set of sample correlation coefficients measured against a known set of parameters.
Yes, you can rank them. Their logical ranking is against a utility function. Would there ever be a reason to rank a set? Yes, for demonstration.
If you were teaching the idea of representativeness, you could use the plot of the raw data over the various $1\dots{k}$ variables in the most and least representative sample. If $n$ is large, it might give you an idea of how wide that spread could be.
The advantage in the field of statistics created by measuring an estimated set against a known set is that it permits a statistician to discover, or sometimes simulate, a sampling distribution for $\hat{\rho}$ and from that to know a number of essential properties. Those properties then make it into textbooks as the solution to a known problem so that the researcher can simply plug in the best estimator in a circumstance.
A t-test or an F-test is used because of these properties. Alternative tests are used when those properties do not hold. Likewise, the mean, median, mode, variance, interquartile range, and so on are used because of their properties.
This method is not sensible if you do not actually know the true correlation matrix. It is only a sensible decision if you have worked out the underlying statistical properties. If you assume them into existence and nature does not agree with you, you can create incorrect tools for a real-world problem.
With real-world sets, especially with sets that have missing data or begin at different times, various techniques exist to deal with this problem. Your question would need more definition to help much further.
Your question tag contains the phrase, “modern portfolio theory,” which is a Frequentist construction. Is Modern Portfolio Theory essential to your question because it will also determine your loss function if it is? It will also determine your school of thought and impose boundaries on your possible set of probability distributions. Doing so also restricts how you handle missing data and outliers. Imposing Modern Portfolio Theory onto data imposes many methodological restrictions that are not otherwise present in statistical theory.
For example, Modern Portfolio Theory does not hold in a Bayesian set of assumptions. Indeed, it is wholly foreclosed under most axiomatic constructions because countable additivity is excluded. You cannot create partitions with an infinite number of sets. However, missing data can be solved in a different manner than it can be solved in non-Bayesian methods.
Bayesian methods do not necessarily distinguish between parameters and data. Instead, the discussion tends to be about observables and unobservables. A missing piece of data that was collected but accidentally erased and lost is now unobservable as may be a population mean. On the other hand, a parameter may be known in some cases. It most commonly happens in an engineering setting. In that case, the known parameter is treated in Bayesian methods just like it is observed data because it is known.
Imposing Modern Portfolio Theory also imposes rules on how to handle missing data. It chooses your estimators. It determines your tests.
That is a wonderful thing in some ways. If the theory is strictly correct, then all kinds of decisions have been made for you that you do not have to think about. If it isn’t, then the field of statistics offers a pretty robust set of tools. Still, now you have to think about your real utility function, the actual distributions involved, and possibly what you may need as your set of probability axioms.
If you want to say that 2013 was the year that had observations that were the best compared to a true value for the population parameters, then the answer is “no.” If you want to know the properties of some choice of estimator construction, then a ranking isn’t necessarily the best tool, but it would be if you were demonstrating something.
It is important to remember that tossing a fair coin so that it comes up heads ten times in a row is consistent with a fair coin over a large enough number of samples. If the first sample happens to have ten heads in a row, that is consistent with a fair coin. It is unusual but consistent.
