I claim that under these assumptions, the functions $f_n$ are uniformly bounded. Then the conclusion follows from the dominated convergence theorem.

First set $H(x) = \int_0^x \int_0^t h(s)\,ds$, which is $C^1$. Letting $u_n = f_n-H$, we have that $u_n$ is concave (i.e. $-u_n$ is convex), continuous, and still converges pointwise.

Now let $v_n(x) = u_n(x) - (x u_n(1) + (1-x) u_n(0))$. Now $v_n$ is again concave and continuous, $v_n(0)= v_n(1)=0$, and $v_n$ still converges pointwise (note that $u_n(0), u_n(1)$ both converge to finite limits). In particular we have $v_n \ge 0$ everywhere.

Let $M_n$ be the maximum value of $v_n$, and let $x_n$ be the point where it is attained. Suppose first that $x_n \ge 1/2$. By concavity we have $v_n(x) \ge \frac{x}{x_n} v(x_n) \ge x v(x_n) = x M_n$ for all $0 \le x \le x_n$. In particular, we have $v_n(1/2) \ge \frac{1}{2} M_n$. If $x_n \le 1/2$, we can get the same result by a similar argument (or by replacing $v_n(x)$ with $v_n(1-x)$).

So $M := \sup_n M_n \le 2 \sup_n v_n(1/2)$ which is finite because $v_n$ converges pointwise. So we have $0 \le v_n(x) \le M$ for all $x,n$. It follows easily that $f_n$ is uniformly bounded as well (by, say, $M + \sup_n |u_n(0)| + \sup_n |u_n(1)| + \sup_x |H(x)|$).