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 $$C_{G'}(s')\supsetneqq T'.$$