Theoretically: no.
For most practical purposes: yes; given that risks are small risks, see these lecture notes on p76.
Belows's the background and one example showing you why you can run into problems:
Given two portfolio compositions $P_i,P_j$ with $(\mu_i,\sigma_i)$ and $(\mu_j,\sigma_j)$, the agent prefers $i$ over $j$ if
$E(u(P_i))\geq E(u(P_j))$.
In your example, the Taylor approximation reads
$$\begin{align}
E\left[u(1+x)\right]&=E\left[u(1+\mu+x-\mu)\right]\\
&=E\left[u(1+\mu)+\sum_{i=1}^{\infty}\left.\frac{\partial^k u}{\partial x^k}\right|_{x=(1+\mu)}\frac{(x-\mu)^k}{i!}\right]\\
&\approx u(1+\mu)+\frac{1}{2}u''(1+\mu)\sigma^2\\
&\equiv \tilde{E}_2[u(1+x)]
\end{align}
$$
This approximation holds well in most cases, but it will break in corner cases as higher order terms are lost in the approximation. Note that each contribution from the higher order terms is negative; thus we can have $E_2(u)$ induce a different ordering of investment opportunities than $E(u)$, as $E_2(u)$ does not consider all higher order effects.
Here's an (somewhat contrived) example: Given the two normal distributions in the table below, we see that the second order Taylor approximation (E2) yields a preference for distribution 2 over 1, the same holds when increasing the length of the Taylor approximation up to the sixth order (E4,E6). Only after incorporating the eighth order (or more), we see that the Taylor approximation reveals the true preference relation, i.e. that distribution 1 is preferred over 2.
i mu sigma E2 E4 E6 E8 .. E(u)
1 0.0795066 0.10 - - - + .. +
2 0.0800000 0.11 + + + - .. -
For most practical purposes, this error should be sufficiently small. Eg when comparing investment choices, the error bound should be so tight that you can work with approximate results.
HTH?