I think with extra assumption $\int f(u_{k}) \to \int f(u)$ one can prove weak convergence. Crucial observation, already made by Pietro Majer, is that $f(u \chi_{A}) = f(u) \chi_{A}$, since $f(0)=0$. If I recall correctly, convex functions are lower semicontinuous with respect to weak topology (it should be easily verifiable), so our assumptions give
$$ \int f(u) \chi_{A} \leqslant \liminf_{n\to \infty} \int f(u_{n}) \chi_{A}.$$
But we can apply exactly the same reasoning to $\chi_{A'}$ to arrive at

$$ \int f(u) \leqslant \liminf_{n \to \infty} \int f(u_n) \chi_{A} + \liminf_{n \to \infty} \int f(u_n) \chi_{A'} \leqslant \liminf \int f(u_n) = \int f(u).$$
Hence,

$$ \liminf \int f(u_n) \chi_A = \int f(u) \chi_A$$
$$ \liminf \int f(u_n) \chi_{A'} = \int f(u) \chi_{A'}$$

What is more, $\liminf \int f(u_n) \chi_{A'} =\lim \int f(u_n) - \limsup \int f(u_n) \chi_{A}$, so $ \lim \int f(u_n) \chi_{A} = \int f(u) \chi_{A}$. Characteristic functions are linearly dense in $L^{\infty}$ and sequence $f(u_n)$ is bounded in $L^1$, so this implies weak convergence.