The answer is yes. The following proof is much less elegant than Jaguar's but has the minor virtue of being self-contained, in that it doesn't refer to the Erdős–Dushnik–Miller theorem. Mainly, though, I'm posting it because I lay awake in bed most of last night working it out and I hate to see that misery go to waste.
Theorem. Let $A$ be an index set, $|A|=\aleph_1$. Let $I_n$ ($n\in\mathbb N$) be disjoint countable sets with $I=\bigcup_{n\in\mathbb N}I_n$, and let $\mu$ be a nonnegative finitely additive measure on $I$ such that $\mu(I_n)\lt\infty$ for all $n\in\mathbb N$. Let $\mathcal U$ be a free ultrafilter on $\mathbb N$ and let $\varepsilon\gt0$. For each $\alpha\in A$ let $S_\alpha$ be a set such that
$$\{n\in\mathbb N:\mu(S_\alpha\cap I_n)\gt\varepsilon\mu(I_n)\}\in\mathcal U$$
(so in particular $|S_\alpha\cap I|=\aleph_0$).
Then there is an infinite set $M\subseteq A$ such that $(\bigcap_{\alpha\in M}S_\alpha)\cap I_n\ne\varnothing$ for infinitely many $n\in\mathbb N$.
Proof.
Claim 1. There exist $n_0\in\mathbb N$, $x_0\in I_{n_0}$, $B\subseteq A$, $J_n\subseteq I_n$ ($n\in\mathbb N$), and $J\subseteq I$ such that $|B|=\aleph_1$, $J=\bigcup_{n\in\mathbb N}J_n$,
$$\forall\alpha\in B,\ \{n\in\mathbb N:\mu(S_\alpha\cap J_n)\gt\varepsilon\mu(J_n)\}\in\mathcal U,$$
$x_0\in S_\alpha$ for all $\alpha\in B$, and $I_{n_0}\cap J=\varnothing.$
Proof of Claim 1. Choose $x_0\in I$ so that $\{\alpha\in A:x_0\in S_\alpha\}$ is uncountable. (If there were no such $x_0$ then, since $I$ is countable and $A$ is uncountable, there would be some $\alpha\in A$ with $S_\alpha\cap I=\varnothing$.) Then $x_0\in I_{n_0}$ for some $n_0\in\mathbb N$. Let $B=\{\alpha\in A:x_0\in S_\alpha\}$, let $J_n=I_n\setminus I_{n_0}$ for $n\in\mathbb N$, and let $J=I\setminus I_{n_0}$.
Claim 2. There exist $\alpha_0\in A$, $B\subseteq A\setminus\{\alpha_0\}$, $J_n\subseteq I_n$ ($n\in\mathbb N$), $J\subseteq I$, and $\delta\gt0$ such that $|B|=\aleph_1$, $J=\bigcup_{n\in\mathbb N}J_n\subseteq S_{\alpha_0}$, and
$$\forall\beta\in B,\ \{n\in\mathbb N:\mu(S_\beta\cap J_n)\gt\delta\mu(J_n)\}\in\mathcal U.$$
Proof of Claim 2. Choose $k\in\mathbb N$ so that $k\varepsilon\gt1$ and choose $\delta\gt0$ so that $k\varepsilon-\binom k2\delta\gt1$. Let's call $S_\alpha$ and $S_\beta$ "almost disjoint" if $\{n\in\mathbb N:\mu(S_\alpha\cap S_\beta\cap I_n)\le\delta\mu(I_n)\}\in\mathcal U$. Then any pairwise almost disjoint subfamily of $\{S_\alpha:\alpha\in A\}$ has fewer than $k$ elements, so there is a maximal pairwise almost disjoint subfamily $\{S_\alpha:\alpha\in A_0\}$ where $A_0$ is a finite subset of $A$. Then for each $\beta\in A\setminus A_0$ there is some $\alpha\in A_0$ such that $\{n\in\mathbb N:\mu(S_\beta\cap S_\alpha\cap I_n)\gt\delta\mu(I_n)\}\in\mathcal U$. Choose $\alpha_0\in A_0$ and $B\subseteq A\setminus A_0$ so that $|B|=\aleph_1$ and
$$\forall\beta\in B,\ \{n\in\mathbb N:\mu(S_\beta\cap S_{\alpha_0}\cap I_n)\gt\delta\mu(I_n)\ge\delta\mu(S_{\alpha_0}\cap I_n)\}\in\mathcal U.$$
Let $J_n=S_{\alpha_0}\cap I_n$ ($n\in\mathbb N$), and let $J=S_{\alpha_0}\cap I$.
Applying Claims 1 and 2 infinitely many times, we get infinitely many distinct elements $\alpha_0,\alpha_1,\dots\in A$ and $x_0,x_1,\dots\in I$ and $n_0,n_1,\dots\in\mathbb N$ such that $x_k\in S_j\cap I_{n_k}$ for all $j,k\in\mathbb N$.
P.S. The following is a partial answer to a question raised in a comment.
Theorem. Let $(I_n:n\in\mathbb N)$ be a sequence of disjoint nonempty sets with $|I_n|\le2^{\aleph_0}$. Assuming the continuum hypothesis $2^{\aleph_0}=\aleph_1$, there is a family of sets $(A_t:t\in\mathbb R)$ such that:
(i) $|I_n\setminus A_t|=1$ for all $t\in\mathbb R$ and all $n\in\mathbb N$;
(ii) there is no set $X$ such that $\{n\in\mathbb N:X\cap I_n\ne\varnothing\}$ is infinite and $\{t\in\mathbb R:X\subseteq A_t\}$ is uncountable.
Proof. Let $\{X_s:s\in\mathbb R\}$ be the collection of all countable sets $X\subseteq\bigcup_{n\in\mathbb N}I_n$ such that $\{n\in\mathbb N:X\cap I_n\ne\varnothing\}$ is infinite.
By the continuum hypothesis, there is a well-ordering $\prec$ of $\mathbb R$ such that, for each $t\in\mathbb R$, the set $\{s\in\mathbb R:s\prec t\}$ is countable.
By transfinite recursion we can define $A_t\subseteq\bigcup_{n\in\mathbb N}I_n$ for $t\in\mathbb R$ so that $|I_n\setminus A_t|=1$ for each $n\in\mathbb N$ while $X_s\not\subseteq A_t$ for all $s\prec t$.