The limit is not $u(\int_0^1 x p(x)\,dx)$ but rather $\int_0^1 u(x) p(x)\,dx$.

I prefer probabilistic notation, so let $X_n \sim p_n$ and $X \sim p$. We are then supposing that $X_n \Rightarrow X$ in distribution. Let $Y_n$ have conditional distribution $\mathrm{Bin}(n, X_n)$ given $X_n$. The question is then about $\lim_{n \to \infty} E[u(Y_n / n)]$. Your conjecture is that it equals $u(E[X])$ but it actually equals $E[u(X)]$.

Fix $\epsilon > 0$. By uniform continuity there exists $\delta > 0$ such that if $|s-t| < \delta$ then $|u(s)-u(t) < \epsilon|$. Now by Chebyshev's inequality, we have
$$\begin{align*} P\left(\left|\frac{Y_n}{n} - X_n\right| \ge \delta\right) &\le \delta^{-2} \operatorname{Var}\left( \frac{Y_n}{n} - X_n\right) \\
&= \delta^{-2} E\left[\operatorname{Var}\left(\frac{Y_n}{n} \mid X_n \right)\right] \\
&= \delta^{-2} n^{-1} E[X_n (1-X_n)] \\
&\le \delta^{-2} n^{-1}.\end{align*}$$
Hence
$P\left(|u(\frac{Y_n}{n}) - u(X_n)| \ge \epsilon\right) \le \delta^{-2} n^{-1}$, so $|u(\frac{Y_n}{n}) - u(X_n)| \to 0$ in probability. By dominated convergence (since $u$ is bounded), $E[u(\frac{Y_n}{n})] - E[u(X_n)] \to 0$ also. But since $X_n \Rightarrow X$, we have $E[u(X_n)] \to E[u(X)]$.