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