The answer is yes. Indeed, let $u_n:=f_n-H$ and $v:=g-H$, where \begin{equation} H(x):=\int_0^x(x-t)h(t)\,dt \end{equation} for $x\in[0,1]$. Then $u_n\to v$ pointwise and $u_n$ is concave for each $n$ (see the Detail below). Hence, $v$ is concave, and so, $u_n$ and $v$ are continuous and piecewise-monotone. So (see e.g. https://math.stackexchange.com/questions/834126/sequence-of-monotone-functions-converging-to-a-continuous-limit-is-the-converge), $u_n\to v$ uniformly and hence $f_n\to g$ uniformly, which implies $\lim \int_0^1 f_n=\int_0^1 g$.
Detail: By Taylor's theorem with the integral form of the remainder and the definition of $H$, for $x\in[0,1]$,
\begin{equation}
u_n(x)=f_n(x)-H(x)=f_n(0)+f'_n(0)x+\int_0^1(x-t)_+[f''_n(t)-h(t)]\,dt.
\end{equation}
Now the concavity of $u_n$ follows because $(x-t)_+$ is convex in $x$ and $f''_n<h$.