First, in order to avoid Amit's concern, we may as well assume countable choice. Note that if AD (axiom of determinacy) holds in $L(\mathbb{R})$, then DC (axiom of dependent choice) will also be true there, and DC implies countable choice.
What you are asking for then is related to Jech's notion of stationarity. Specifically, we say that $S \subseteq [A]^{\omega}$ ($[A]^{\omega}$ is the set of countable subsets of $A$) is stationary when $S$ meets every club, where club here is in the sense you described. Now you can verify that the club filter on $[A]^{\omega}$ is an ultrafilter if and only if $[A]^{\omega}$ cannot be decomposed into two disjoint stationary sets.
Although not true for arbitrary $A$, you can also verify that $S \subseteq [\omega_1]^{\omega}$ is stationary according to Jech's characterization if and only if $S \cap \omega_1$ is stationary in the usual sense. Since AD implies that the club filter on $\omega_1$ is an ultrafilter ( Model of ZF + $\neg$C in which Solovay's Theorem on stationary sets fails? ), I claim that the club filter on $[\omega_1]^{\omega}$ is actually an ultrafilter if AD is true for suppose not. Then the club filter on $[\omega_1]^{\omega}$ could be decomposed into two disjoint stationary sets $S_1$ and $S_2 = [\omega_1]^{\omega} \setminus S_1$ so that $S_1 \cap \omega_1$ and $S_2 \cap \omega_1$ would be disjoint stationary sets of $\omega_1$. But this is impossible because then $S_1 \cap \omega_1$ and its complement would not be in the club filter on $\omega_1$.
