Not if $n\gt 1$. Consider the subring of all scalar multiples of the identity matrix (this is isomorphic to $\mathbb{F}$ itself).
Since the collection of all matrices that commute with $A$ always includes $A$ itself, if the subring of all scalar multiples of the identity were of the form $C(A)$ for some $A$, then $A$ would necessarily be a scalar multiple of the identity; but scalar multiples of the identity are central in $\mathbb{F}^{n\times n}$, so the centralizer of such an $A$ would contain more than just the scalar multiples of the identity.
You might ask more generally whether every subring of $\mathbb{F}^{n\times n}$ of the form $C(S)$ for some $S\subseteq \mathbb{F}^{n\times n}$, where $C(S) = \{M\in\mathbb{F}^{n\times n}\mid MA=AM\text{ for all }A\in S\}$. But even here the answer is still negative; though my example no longer works in this setting (taking $S=\mathbb{F}^{n\times n}$ gives the scalar multiples of the identity), Mariano's example still does. Note that $S\subseteq C(C(S))$; if $T$ is the set of upper triangular matrices, then considering the matrix $E_{ij}$ that has a $1$ in the $(i,j)$ entry, $i\leq j$, and zeros elsewhere, you have that $E_{ij}A$ is the matrix that has zeros everywhere except the $i$th row, where it has the $j$th row of $A$; while $AE_{ij}$ is the matrix that has zeros everywhere except the $j$th column, where it has the $i$th column of $A$. Thus, if $A\in C(T)$, then $A$ must be a scalar multiple of the identity; this means that if $T=C(S)$ for some $S$, then $S$ must be contained in the scalar multiples of the identity, and once again we obtain a contradiction.