This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theore This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (by Glauberman) without CFSG for $p =2,$ but as far as I know all proofs to date for odd $p$ use CFSG.
Now suppose that $G$ is a doubly transitive permutation group with $F(G) =1,$ and that the point stabilizer $G_{x}$ has a central element $u$ of prime order $p$ (which will certainly happen if $Z(G_{x}) \neq 1).$ Then $G_{x} = C_{G}(u)$ and the permutation action is equivalent to that of $G$ acting by conjugation on the conjugates of $u.$
Now $C_{G}(u)$ permutes the conjugates of $u$ which commute with $u.$ By the double transitivity of the permutation action, we see that either all conjugates of $u$ commute with $u,$ or else no conjugate of $u$ other than $u$ itself commutes with $u.$ In the former case, $u \in F(G),$ which is excluded by hypothesis. Hence $u$ commutes with none of its other conjugates. By the general $Z^{\ast}$-theorem, we either have $u \in Z(G)$ or else $O_{p^{\prime}}(G) \neq 1.$ The former case is excluded as $F(G) = 1.$
Suppose then that $O_{p^{\prime}}(G) \neq 1.$ Then the image of $u$ is central in $G/O_{p^{\prime}}(G)$ and a Frattini argument yields $G = O_{p^{\prime}}(G)C_{G}(u).$
Since $u \not \in Z(G),$ there is a prime $q \neq p$ such that $u$ normalizes, but does not centralize, a Sylow $q$-subgroup, $Q$ say, of $O_{p^{\prime}}(G)$. Hence there is a $Q$-conjugate $v$ of $u$ such that $\langle u,v \rangle$ is a $\{p,q\}$-group. By the transitivity of $C_{G}(u)$ on the other conjugates of $u,$ it follows that $\langle u,w \rangle$ is a $\{p,q\}$-group for every conjugate $w$ of $u.$ It then follows that $O_{p^{\prime}}(G)$ is a $q$-group, and that $F(G) \neq 1,$ contrary to hypothesis.