The answer is no. Take $F$ to be reals, and consider the subgroup $K=U(g)\subset Sp_{2g}(F)$. Then $K$ is the centraliser of "multiplication by $i$", and is not quasi-split since it is compact. The element multiplication by $i$ on $\mathbb{C}^g$ is to be viewed as the $g$-fold direct sum of the two by two matrix $\begin{pmatrix} 0 & 1 \cr -1 & 0\end{pmatrix}$ on the real vector space $(\mathbb{R}^2)^g$.

A similar example can be given over non-archimedean local fields. The resulting group will not be anisotropic over $F$, but will not be quasi-split. To see this, let $E/F$ be a quadratic extension and denote by $x\mapsto \overline{x}$ the action by the non-trivial element of the Galois group. Let $h$ be a non-degenerate Hermitian form with respect to $E/F$ on the $E$ vector space $E^g$. The relative norm one elements $S$ of $E/F$ lie in $U(h)$. The "imaginary part" $\Omega$ of $h$ is a non-degenerate symplectic form on $F^{2g}=E^g$ and the centraliser of a non-zero element $X\in Lie (S)\subset Lie Sp_{2g}$ is precisely $U(h)$. For a suitable choice of $h$, the group $U(h)$ is not quasi split. (If the number of variables $g\geq 3$, the Hermitian form does represent a zero for non-archimedean local fields, so the group $U(h)$ is isotropic).