Lyubomyr Zdomskyy informed me that the answer to my question is affirmative and follows from a recent (still unpublished) result of Osvaldo Guzman and Damjan Kalajdzievski who proved the consistency of $\mathfrak u<\mathfrak a=\mathfrak c$.
Take any ultrafilter $\mathcal U$ on $\omega$ that has a base $\mathcal B\subseteq\mathcal U$ of cardinality $|\mathcal B|=\mathfrak u$. Next, choose a maximal almost disjoint family of sets $\mathcal A\subseteq [\omega]^\omega\setminus\mathcal U$. Consider the filter $$\mathcal F=\{F\subseteq \omega:|\{A\in\mathcal A:A\not\subseteq^* F\}|<\mathfrak c\}.$$ We claim that $\mathcal F\subseteq\mathcal U$ under $\mathfrak a=\mathfrak c$. Assuming that $\mathcal F\not\subseteq\mathcal U$, we can find a set $F\in\mathcal F\setminus\mathcal U$. For the set $F$ the family $\mathcal A'=\{A\in\mathcal A:A\not\subseteq^* F\}$ has cardinality $|\mathcal A'|<\mathfrak c=\mathfrak a$. Then the family $\{A\setminus F\}_{A\in\mathcal A'}$ is not maximal almost disjoint in $\omega\setminus F\in\mathcal U$. So, we can choose an infinite set $B\subseteq\omega\setminus F$ such that $B\notin \mathcal A$ and the family $\{B\}\cup\{A\setminus F\}_{A\in\mathcal A'}$ is almost disjoint. Replacing $B$ by a smaller infinite subset we can additionally assume that $B\notin\mathcal U$. Then $\{B\}\cup\mathcal A$ is an almost disjoint family in $[\omega]^\omega\setminus\mathcal U$, which is strictly larger than $\mathcal A$. But this contradicts the maximality of $\mathcal A$. This contradiction shows that $\mathcal F\subseteq\mathcal U$.
