It should be noted that, in general, for any non-linear function $g$ of a random variable $X$,
$
\mathbb{E}[g(X)] \neq g(\mathbb{E}[X])
$
. You can see this by expanding out the integrals;
$$
\mathbb{E}[g(X)] = \int_X g(x)f(x)dx;\;\;\;\;g(\mathbb{E}[X])=g\bigg(\int_X xf(x)dx\bigg)
$$
You could see how the two would only be equal in the case $g$ was linear.
So given the above;
$$\mathbb{E}_{\beta}\left[\log \Phi \left(x^{T}\beta\right)\right] \neq \log \Phi\left(x^{T}\mu\right)$$
You can estimate $\mathbb{E}_{\beta}\left[\log \Phi \left(x^{T}\beta\right)\right] $ via Monte Carlo
$$
\mathbb{E}_{\beta}\left[\log \Phi \left(x^{T}\beta\right)\right]=\int_\beta \log \Phi \left(x^{T}\beta\right)f(\beta|\mu, \Sigma)d\beta \approx \frac{1}{G}\sum_{g=1}^G \log \Phi \left(x^{T}\beta^{(g)}\right)
$$
where $\beta^{(1)},\beta^{(2)},...,\beta^{(G)} \sim \mathcal{N}(\mu,\Sigma)$.
If the dimension of $\beta$ is large, the above approximation may not work efficiently (it will take a very large $G$ for it to get close). Otherwise, this is a pretty standard and well accepted approach.
I am not aware of a closed form solution to the above integral. If another user does derive an analytical solution you should use that instead of the the Monte Carlo approximation above. There may also be ways to improve the above estimate with importance sampling.