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 (see the Detail below). Hence, $v$ is concave, and so, $v$ is 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 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 $f''_n<h$.