Let $\Omega\subset\mathbb{R}^n$ be open bounded, let $I\subset\mathbb{R}$ be a compact interval, and let $1\leq p <\infty$. Suppose we have a function $F:I\times\Omega \to \mathbb{R}$ with the following properties:
$\bullet$ $F(s,\cdot)\in L^p(\Omega)$ for every $s\in I$. Here, the underlying measure is the Lebesgue measure on $\Omega$.
$\bullet$ The function $s \mapsto F(s,\cdot)$ is continuous as a function $I\to L^p(\Omega)$.
Is $F$ measurable with respect to the product Borel (or Lebesgue) $\sigma$-algebra on $I\times\Omega$?
Your assumptions imply that $F \in L^\infty (I, L^p(\Omega)) \subset L^p (I, L^p(\Omega))$. Thus, the argument used here Is $L^2(0,\infty;L^2(\Omega)) = L^2((0,\infty)\times \Omega)$? shows that $F$ has a representative (w.r.t. equality almost everywhere) which is in $L^2(I \times \Omega)$ and thus in particular measurable.
As long as you use the Lebesgue $\sigma$-algebra, this means that $F$ itself is measurable.
For the Borel $\sigma$-Algebra, we can just take $F(x,y) = \chi_{\Bbb{V}} (x) \cdot \chi_{\Bbb{Q}}(y)$, where $\Bbb{V}$ is nonmeasurable to construct a non Borel-measurable function which satisfies $F(x, \cdot) = 0$ as an $L^p$-function for all $x$. Hence, your requirements are fulfilled, but $F$ is not Borel measurable.