**Edit**: Previous answer was bogus.

Let me change your notation a little to let $x_0$ be the limit of the $x_n$.

Yes, this is true. It follows from tightness and the general fact that $h(x_n, \cdot) \to h(x_0, \cdot)$ uniformly on compact sets.

Without loss of generality, let's suppose that $h(x_0, \cdot) = 0$ (replace $h(x,y)$ by $h(x,y) - h(x_0,y)$).

Let $\epsilon > 0$ and let $M$ be the sup norm of $h$. By the easier direction of Prohorov's theorem, the sequence $\{\mu_n\}$ is tight, so there is a compact $K \subset S$ such that for every $n$ we have $\mu_n(K^C) < \epsilon$. Now we have
$$\begin{align*}\left|\int h(x_n, y) \mu_n(dy)\right| &\le \int_K |h(x_n,y)|\,\mu_n(dy) + \int_{K^C} |h(x_n, y)| \mu_n(dy) \\
&\le \sup_{y \in K} |h(x_n, y)| + M \epsilon.\end{align*}$$
So it suffices to show that $h(x_n, \cdot) \to 0$ uniformly on $K$. Suppose not; then passing to a subsequence if necessary, we can find $\delta > 0$ so that $\sup_{y \in K} |h(x_n, y)| > \delta$ for all $n$. Thus for each $n$ we can find $y_n \in K$ such that $|h(x_n, y_n)| > \delta$. By compactness, we can pass to a further subsequence so that $y_n$ converges to some $y_0$. Now by continuity of $h$ we have $|h(x_n, y_n)| \to |h(x_0, y_0)| = 0$, a contradiction.