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 $G={\rm GL}_{2,L}$, $G'=G/\{\pm 1\}$.
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 
$$
n=\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}.
$$
Then
$$ n s n^{-1} = -s.$$
This means that if we denote by $s'$ and $n'$ 
the images in $G'$ of $s$ and $n$, respectively, then
$$ n' s' (n')^{-1} = s'.$$
Thus 
$$ n'\in C_{G'}(s'),$$
but $n'\notin T'$, where $T'$ denotes the image of $T$ in $G'$.
We see that  $$C_{G'}(s')\supsetneqq T'.$$