I think I can even manage to prove strong convergence in $L^{1}$. I'm going to show that every subsequence has a further subsequence which converges. So, passing to subsequence, we may assume that $u_{k} \to u$ almost everywhere. Since convex functions are continuous (maybe not at 0), we have convergence $f(u_{k}) \to f(u)$ almost evyrewhere. With assumption $\int f(u_{k}) \to \int f(u)$, we can use Scheffe's theorem which states that if a sequence of densities converges almost everywhere to a density, then it is, in fact, strong convergence in $L^{1}$. Of course, we only have $\int f(u_{k}) \to \int f(u)$ instead of $\int f(u_{k}) = \int f(u)$ but it requires no modification of standard proof.

I hope it's quite clear and, what is even more important, doesn't contain any mistake.

I think I can even manage to prove strong convergence in $L^{1}$. I'm going to show that every subsequence has a further subsequence which converges. So, passing to subsequence, we may assume that $u_{k} \to u$ almost everywhere. Since convex functions are continuous (maybe not at $0$)*, we have convergence $f(u_{k}) \to f(u)$ almost evyrewhere. With assumption $\int f(u_{k}) \to \int f(u)$, we can use Scheffe's theorem which states that if a sequence of densities converges almost everywhere to a density, then it is, in fact, strong convergence in $L^{1}$. Of course, we only have $\int f(u_{k}) \to \int f(u)$ instead of $\int f(u_{k}) = \int f(u)$ but it requires no modification of standard proof.

I hope it's quite clear and, what is even more important, doesn't contain any mistake.

*Since $f(0)=0$ and $f$ is nonnegative and convex, then it is also continuous at $0$.

As commented by Pietro Majer, only continuity of $f$ really matters.