Replacing missing value by class conditional mean
Your intuition about 'no effect' is true in some sense. But this replacement may be not the best use of the information you have. 

The choice of missing value treatment depends on your initial problem statement. In all the cases I assume that you have already somehow estimated conditional means $\mu_0$ and $\mu_1$ and the common variance matrix $S$.

Problem 1. For your test observation, you know $y$, but know nothing about $x$, and want to estimate $x_n$. If this is the case, $\mu_{y, n}$ is indeed your best guess of $x_n$. 

Problem 2. For your test observation, you know $x_{-n}$ (that is, $x$ except $x_n$) and $y$, and want to predict $x_n$ as best as possible. If it is the case, you should use expectation conditional not only on $y$, but also on $x_{-n}$:
$$
\hat{x_n} = \mathbb{E}(x_n|y, x_{-n}) = \mu_{(y,n)} + S_{(n,-n)}S_{(-n)}^{-1}(x_{-n}-\mu_{(y,-n)})
$$ 
where $S_{*}$, $\mu_{y,*}$ are corresponding parts of covariance matrix and conditional expectation vector.

Problem 3. You know $x_{-n}$ but don't know $y$, and you want to estimate $x_n$. In this case, conditional distribution of $x_n$ is a mixture of two normal distributions, $p(x_n|y=0, x_{-n})$ and $p(x_n|y=1, x_{-n})$, with weights, proportional to $p(y=0|x_{-n})$ and $p(y=1|x_{-n})$ respectively. The mean of this distribution is just the weighted average of two means conditional on $y$. But this distribution may be bimodal, and its mean may be not a very useful statistic. 

Problem 4. You know only $x_{-n}$, and your final goal is to estimate probability of $y=1$. In this case, you should just forget about $x_n$ (you don't know anything of it anyway), and use the distribution of $x_{-n}$ conditional on $y$ for whatever algorithm you apply.