Since the question has now changed multiple times, I am editing in a new answer after some discussion in the comments.

**New question/answer**:

Let $X$ be a separable and reflexive Banach space. Let $u_n$ be continuous functions $[0,T] \to X$ which are uniformly bounded, so there exists $C > 0$ such that $\|u_n(t)\|_X \leq C$ for all $t \in [0,T]$ and all $n$. Suppose that $(u_n)$ converges to some $u$ weakly in $L^2(0,T;X)$. If $F \colon X \to X$ is locally Lipschitz and weakly continuous on $X$, does $(F(u_n))$ converge weakly to $F(u)$ in $L^2(0,T;X)$?

This is wrong in general. Consider $L^2(0,2\pi)$, so $T = 2\pi$ and $X = \mathbb{R}$. Set $F(x) = x^2$ as a real function. It is locally Lipschitz continuous on $\mathbb{R}$, in particular weakly continuous.

The functions $$u_n(t) := \frac1{\sqrt{\pi}} \sin \left(\frac{nt}2\right)$$ are continuous and uniformly bounded. They form an orthonormal basis in $L^2(0,2\pi)$, thus they converge weakly to $u = 0$ there. But testing $F(u_n)$ against the constant $1$ function $\mathbf{1} \in L^2(0,2\pi)$ gives $$1 = \|u_n\|^2_{L^2(0,2\pi)} = \int_0^{2\pi} u_n(t)^2 \, dt = \int_0^{2\pi} F(u_n(t)) \, dt = \bigl(F(u_n),\mathbf{1}\bigr)_{L^2(0,2\pi)}$$
for all $n$, hence $F(u_n)$ does not converge weakly to $F(u) = 0$ in $L^2(0,2\pi)$.

**Old question/answer**:

Let $X$ be a separable and reflexive Banach space. Suppose that $u_n$ converges to some $u$ weakly in $L^2(0,T;X)$. If $F \colon X \to X$ is locally Lipschitz and weakly continuous on $X$, does $F(u_n)$ converge weakly to $F(u)$ in $L^2(0,T;X)$?

Weakly continuous Nemytskii operators on Lebesgue (or function) spaces are a rare breed and this is already the case for $X = \mathbb{C}$ or $\mathbb{R}$. More specifically, a weakly continuous Nemytskii operator mapping, say, $L^p(0,1)$ into itself is necessarily already affine linear. See this question and the answers there.

**Old query by myself to a remark in the question**:

As a sidenote, I do not quite understand how you get $F(u_n(t))$ that converges weakly to $F(u(t))$ in $X$ for (almost?) every $t \in (0,T)$ from your assumptions, maybe there is a misunderstanding here?

This claim is also false as shown in the comments to the OP, using the ONB in $L^2(0,2\pi)$ and $F(x) = x$.