$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\LL}{\mathcal L}\newcommand{\si}{\sigma}$The answer is yes, at least for $p\in[1,\infty)$.
Indeed, $L^p(\R^d)$ is a separable metric 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, $\|G_\ep(t,\cdot)-F(t)\|_{L^p(\R^d)}\le\ep$ and hence $\|G_\ep(t,\cdot)\|_{L^p(\R^d)}\le\ep+\|F(t)\|_{L^p(\R^d)}$ for each $t\in[0,T]$.
Since $F\colon[0,T]\to L^p(\R^d)$ is measurable and the norm on $L^p(\R^d)$ is continuous and hence measurable, we see that the function $[0,T]\ni t\mapsto w(t):=\dfrac1{1+\|F(t)\|^p_{L^p(\R^d)}}\in[0,\infty)$ is measurable. So, for each real $\ep>0$ we have $G_\ep\in L^p_w([0,T]\times\R^d)$, where $L^p_w([0,T]\times\R^d)$ is the space of all measurable functions $H\colon[0,T]\times\R^d\to\R$ with norm
$$\|H\|_{L^p_w([0,T]\times\R^d)}:=\Big(\int_0^T dt\,w(t)\,\|H(t,\cdot)\|_{L^p(\R^d)}^p\Big)^{1/p}<\infty.$$
For all integers $m,n$ such that $m\ge n\ge1$
\begin{equation}
\|G_{1/m}-G_{1/n}\|_{L^p_w([0,T]\times\R^d)}^p
=\int_0^T dt\,w(t)\|G_{1/m}(t,\cdot)-G_{1/n}(t,\cdot)\|_{L^p(\R^d)}^p\le(2/n)^pT\to0
\end{equation}
as $n\to\infty$. So, by the completeness of $L^p_w([0,T]\times\R^d)$, for some sequence $(n_k)$ of natural numbers going to $\infty$ there is a limit $G$ of $G_{1/n_k}$ in $L^p_w([0,T]\times\R^d)$. Clearly, this limit $G$ satisfies your desired conditions. $\quad\Box$
Best Answer
Sure. Triangulate the unit cube $[0,1]^d$ into $d!$ simplices $S_1$, $\ldots$, $S_{d!}$, as in any of the answers to this question.
Let $g$ satisfy $g(\vec{n}) = f(\vec{n})$ for all $\vec{n} \in \mathbb{Z}^d$, and make it linear on $S_i + \vec{n}$ for every $1 \leq i \leq d!$ and $\vec{n} \in \mathbb{Z}$. Then $g$ is Lipschitz and $f - g$ is bounded.