Note:
It is computationally simple to determine the volatility of any given return series, so in fact there may be no need for this approximation.
Let's start with the annualized return $r_a$, which is
$$r_a = \sqrt[T]{1+R_t}-1$$
where $R_t$ is the cumulative return over the whole period $[0,T]$. Consider the Taylor-approximation
$$log(1+y) = y - \frac{1}{2}y^2 + \frac{1}{3}y^3 - \frac{1}{4}y^4 +...$$
Taking the first two terms, you obtain:
$$log(1+r_a) \approx r_a - \frac{1}{2}\sigma_{r_a}^2$$
However, if $r_a$ is not approximately zero, the error becomes great. The following image shows this error $log(1+y) - y$ within the interval $y \in [-0.8,0.8]$:

If any annualized return $r_a$ is smaller than -39% or greater than 52%, the error from the approximation exceeds 10%!
Further, most financial asset returns have negative skewness and leptokurtosis, so the approximation above is biased upwards. In fact, you may adjust for this and use the formula
$$log(1+r_a) \approx r_a - k\frac{1}{2}\sigma_{r_a}^2$$
,where $k$ is an empirical factor (often between 5 and 10), see here.
Re-arranging gives you an approximation for the volatility of the return series:
$$\sigma_{r_a} \approx \sqrt{\frac{2r_a - 2log(1+r_a)}{k}}$$
EDIT
OP is asking on how to fit a curve of $\sigma(R_T)$ in terms of $T$. Let me provide you the results of a simulation run in R. I use 100,000 daily returns following a normal distribution $N(0.01/252, 0.005)$, so i assume a mean return of one percent per year with a standard deviation of 7,94% ($0.005 \cdot \sqrt{252}$):
set.seed(100)
r = rnorm(100000, .01/252, .005)
## Vector containing the cumulative simple return up to T
cum <- vector(mode = "numeric", length = 100000)
## Vector containing the volatility up to T
vola <- vector(mode = "numeric", length = 100000)
for(i in 1:100000){cum[i] <- prod(1+r[1:i])}
for(i in 1:100000){vola[i] <- sd(cum[1:i])}
## vola[1] is NA due to vola[1] <- sd(cum[1:1]),
## so we set it to zero
vola[1] <- 0
summary(cum)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.8405 1.2581 2.1961 5.1116 5.0303 36.2721
summary(vola)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0000 0.1775 0.4096 1.1516 1.1164 6.3274
The following image shows the variable vola
, i.e. $\sigma (R_T)$ for $T \in [0; 100,000]$:

We see, that $\sigma (R_T)$ is non-linear and increasing in time. I implement your suggested regression $$\sigma(R_T)\sim aT^\alpha + b$$ using time $T$ with single steps of $\frac{1}{100,000}$, which results in:
## set up variable time
time <- seq(0, 1, 0.00001)
## eliminate first value of time which is zero,
## so we have single time steps of 1/100,000
time <- time[-1]
reg <- nls(vola ~ a*(time^alpha)+b, start = list(a=1, alpha=1, b=1))
summary(reg)
Parameters:
Estimate Std. Error t value Pr(>|t|)
a 5.914458 0.003736 1582.9 <2e-16 ***
b 0.208985 0.001096 190.7 <2e-16 ***
alpha 5.274693 0.006115 862.6 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.245 on 99997 degrees of freedom
Number of iterations to convergence: 7
Achieved convergence tolerance: 1.396e-06
In summary, your suggested non-linear regression model seems to be useful. However, my starting values were chosen randomly, so you might try to evaluate if other configurations don't affect the results to heavily. You might want to try the glmulti R-package which implements this Automated Model Selection with (Generalized) Linear Models.