Yes.
Let $\mu$ be the Lebesgue measure. We start with
Lemma: function $F$ admits an estimate $$|F(x)| \le C_1|x| + C_2$$ for some constants $C_1, C_2 > 0$.
Proof: Suppose not. Then there is a sequence of points $x_n \to \infty$ such that $$\frac{|F(x_n|)}{|x_n|} \to \infty.$$ Consider a function $u$ which takes value $x_1$ on the set of measure $m_1$, value $x_2$ on the set of measure $m_2$, etc. Then $$||u||^2_{L^2} = m_1x_1^2 + m_2x_2^2 + \ldots, \\ ||F(u)||^2_{L^2} = m_1F(x_1)^2 + m_2F(x_2)^2 + \ldots.$$
Now choose $m_i$ in such a way that only the first of these series is converging. Thus, $F(u) \not \in L^2$ which is a contradiction. End proof
Let us now prove that $||F(w_1 + \eta w_2)||_{L^2}$ is continuous at $\eta = 0$.
Take $n \in \mathbb{N}$ and let $A_n = \{x \in \Omega \mid |w_1(x)| \le n, |w_2(x)| \le n\}$. We have $\mu(\Omega \setminus A_n) \to 0$. On the set $A_n$ function $w_1 + \eta w_2$ is bounded by $2n$ for $\eta < 1$. Let $$M = \max\limits_{|y| \le 2n}|F^{\prime}(y)|.$$
We have, using Lemma, $$||F(w_1) - F(w_1 + \eta w_2)||^2_{L^2(\Omega)} = ||F(w_1) - F(w_1 + \eta w_2)||^2_{L^2(A_n)} + \\ ||F(w_1) - F(w_1 + \eta w_2)||^2_{L^2(\Omega\setminus A_n)} \le M^2\eta^2||w_2||^2_{L^2(A_n)} + \\ ||C_1(2|w_1| + \eta|w_2|) + 2C_2||^2_{L^2(\Omega\setminus A_n)}.$$
The second term tends to $0$ as $n \to \infty$ and the first term is continuous in $\eta$.
This shows that the function $\eta \mapsto F(w_1 + \eta w_2)$ is continuous as a function from $\mathbb{R}$ to $L^2(\Omega)$. In particular, $\eta \mapsto ||F(w_1 + \eta w_2)||_{L^2}$ is also continuous.