It is a good question! The answer is NO, see the counter-example below.

Take $p=3$; then $\mathbb F_3=\{0,1,-1\}$.
Write $L=\mathbb F_3(i)$, where $i^2=-1$; then $L\simeq \mathbb F_9$.

Take $$G={\rm GL}_{2,L}\,,\quad G'=G/\{\pm 1\}.$$
Let $T\subset G$ denote the subgroup of diagonal matrices.
Take 
$$ s={\rm diag}(i,-i)\in T(L)\subset G(L).$$
Then the centralizer of $s$ in $G$ is $T$, hence $s$ is a regular semisimple element of $G(L)$.

Write 
$$
n=\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\in G(L).
$$
Then
$$ n s n^{-1} ={\rm diag}(-i,i)= -s.$$
This means that if we denote by $s'$ and $n'$ 
the images in $G'(L)$ of $s$ and $n$, respectively, then
$$ n' s' (n')^{-1} = s'.$$
Thus 
$$ n'\in C_{G'}(s')(L),$$
but $n'\notin T'(L)$, where $T'$ denotes the image of $T$ in $G'$.
We see that  $$C_{G'(L)}(s')\supsetneqq T'(L).$$
In the notation of the question, we obtain that 
$$C_{G^{\prime F}}(s')\supsetneqq T^{\prime F}.$$