$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\LL}{\mathcal L}\newcommand{\si}{\sigma} $The answer is yes, at least for $p\in[1,\infty)$.
Indeed, $L^p(\R^d)$ is a separable metric the space. So, for each real $\ep$ there is a countable measurable partition $(B_{\ep,j})$ of $L^p(\R^d)$ such that for each $j$ we have $B_{\ep,j}\ne\emptyset$ and the diameter of $B_{\ep,j}$ is $\le\ep$. Pick any $y_{\ep,j}$ in $B_{\ep,j}$. For $(t,x)\in[0,T]\times\R^d$, let \begin{equation} G_\ep(t,x):=\sum_j y_{\ep,j}(x)\,1(F(t)\in B_{\ep,j}), \end{equation} where $F(t):=F_t$. Then for any real $c$ \begin{equation} \{(t,x)\in[0,T]\times\R^d\colon G_\ep(t,x)>c\} =\bigcup_j F^{-1}(B_{\ep,j})\times y_{\ep,j}^{-1}((c,\infty)) \in\LL([0,T])\otimes\LL(\R^d), \end{equation} where $\LL(\cdot)$ denotes the Lebesgue $\si$-algebra. So, the function $G_\ep$ is measurable, for each $\ep$.
Also, $\|G_\ep(t,\cdot)-F(t)\|_{L^p(\R^d)}\le\ep$ for each $t\in[0,T]$ and hence for all integers $m,n$ such that $m\ge n\ge1$ \begin{equation} \|G_{1/m}-G_{1/n}\|_{L^p([0,T]\times\R^d)}^p =\int_0^T dt\,\|G_{1/m}(t,\cdot)-G_{1/n}(t,\cdot)\|_{L^p(\R^d)}^p\le(2/n)^pT\to0 \end{equation} as $n\to\infty$. Also, in view of the truncation $F(t)\,1(|F(t)|<N)$, without loss of generality $G_\ep\in L^p([0,T]\times\R^d)$ for each $\ep>0$. So, by the completeness of $L^p([0,T]\times\R^d)$, for some sequence $(n_k)$ of natural numbers going to $\infty$ there is a limit $G$ of $G_{1/n_k}$ in $L^p([0,T]\times\R^d)$. Clearly, this limit $G$ satisfies your desired conditions. $\quad\Box$