The following is due to Alan Dow:
In any model obtained by adding $\aleph_2$ many Cohen reals to a model of $\mathsf{CH}$ the statement is false. We force with $\mathbb{P}=\operatorname{Fn}(\omega_2,2)$ and we let $\dot{\mathcal{I}}$, $\dot{\mathcal{J}}$ and $\dot u$ be $\mathbb{P}$-names such that $\dot{\mathcal{I}}$ and $\dot{\mathcal{J}}$ are forced to be ideals and $\dot u$ is forced to be the unique ultrafilter that extends the two associated regular filters.
Now let $M$ be an elementary substructure of a suitable large $H(\theta)$ that has cardinality $\aleph_1$ and that is closed under $\omega$-sequences. Let $\delta=M\cap\omega_2$ and $\mathbb{P}_M=\operatorname{Fn}(\delta,2)$.
By elementarity the $\mathbb{P}_M$-names $\dot{\mathcal{I}}\cap M$ and $\dot{\mathcal{J}}\cap M$ are forced to be ideals and $\dot u\cap M$ is forced to be the unique ultrafilter that extends the two associated regular filters.
Work in $V[G\cap M]$. We write $\mathcal{I}_M$, $\mathcal{I}_M$ and $u_M$ for the interpretations of $\dot{\mathcal{I}}\cap M$, $\dot{\mathcal{J}}\cap M$ and $\dot u\cap M$ in this model.
Note that every element of $u_M$ meets elements of $\mathcal{I}_M$ and $\mathcal{J}_M$. On the other hand if $\mathcal{I}'$ and $\mathcal{J}'$ are countable subfamilies of $\mathcal{I}_M$ and $\mathcal{J}_M$ respectively then some infinite subset, $x$, of $\omega$ separates the elements of $\mathcal{I}'$ from those of $\mathcal{J}'$. This implies that, without loss of generality, for every countable subfamily $\mathcal{J}'$ of $\mathcal{J}_M$ there is an element $x\in u_M$ such that $x\cap y$ is finite for all $y\in\mathcal{J}'$.
Using this one can construct a sequence $\langle b_\alpha:\alpha<\omega_1\rangle$ in $\mathcal{J}_M$ such that every element of $u_M$ contains all but countably many of the $b_\alpha$.
In $V[G]$ consider the next Cohen real $c$, added by $\operatorname{Fn}([\delta,\delta+\omega,2)$, say and assume, without loss of generality, that $c\notin u$.
There must be a set $Y\subseteq c$ that separates $\lbrace x\cap c:x\in\mathcal{I}\rbrace $ from $\lbrace y\cap c:y\in\mathcal{J}\rbrace $. In $V[G\cap M]$ we take names, $\dot c$ and $\dot Y$, for $c$ and $Y$, note that these are $\operatorname{Fn}(C,2)$-names for some countable set $C$. For every $\alpha<\omega_1$ it is forced that $\dot Y\cap b_\alpha$ is finite; by pigeon-holing there will be one $p\in\operatorname{Fn}(C,2)$ and one $n\in\omega$ such that $p$ forces $\dot Y\cap b_\alpha\subseteq n$ for uncountably many $\alpha$. Still in $V[G\cap M]$ let $Y_p=\lbrace k:(\exists q\le p)(q \Vdash k\in\dot Y)\rbrace $. Then $b_\alpha\cap Y_p\subseteq n$ for uncountably many $\alpha$ so that $Y_p\notin u_M$. On the other hand for every $x\in\mathcal{I}_M$ it is forced that $x\cap\dot c\subseteq \dot Y$ (mod finite); this implies that $x\setminus Y_p$ is finite for all $x\in\mathcal{I}_M$, so that $Y_p\in u_M$.