Let $\mathscr{F}$ be a free ultrafilter on $\mathbf{N}$ containing the even integers, and $(x_1,x_2,\ldots)$ be the $\{0,1\}$-valued sequence such that $x_n=1$ if and only if $n$ is even. Then $(x_n)$ is $\mathscr{F}$-convergent to $1$.
Question. Identifying a sequence $(n_k) \in \{0,1\}^{\mathbf{N}}$ with infinitely many $1$s with a subsequence $(x_{n_k})$, is it true that the set of subsequences $$ \{(n_k) \in \{0,1\}^{\mathbf{N}}: x_{n_k} \text{ is }\mathscr{F}\text{-convergent to }0\} $$ is of first Baire category?
Edit: an equivalent formulation, identifying the set $S$ is strictly increasing sequences in $\mathbf{N}$ with the set of sequences in $\{0,1\}^{\mathbf{N}}$ with infinitely many $1$s:
Question. If $\mathscr{F}$ is a free ultrafilter containing $2\mathbf{N}$, is it true that $\{(n_k) \in S: \{k \in \mathbf{N}: n_k \text{ odd }\}\in \mathscr{F}\}$ is meager?
Let me simplify your problem first: we may replace the set of even natural numbers to the whole $\mathbb{N}$. Moreover, for given $f:\mathbb{N}\to 2$, $f$ is $\mathcal{F}$-convergent to zero iff $f^{-1}[\{0\}]\in \mathcal{F}$. Therefore, our problem is equivalent to asking whether $A_\mathcal{F}^0:=\{f \in 2^\omega \mid f^{-1}[\{0\}]\in \mathcal{F}\}$ is meager.
The answer is negative: observe that $2^\omega$ is a disjoint union of $A_\mathcal{F}^0$ and $A_\mathcal{F}^1:=\{f \in 2^\omega \mid f^{-1}[\{1\}]\in \mathcal{F}\}$. Furthermore, the map $\phi:2^\omega\to 2^\omega$ defined by $$\phi(f)(n)=1-f(n)$$ is a homeomorphism over $2^\omega$ which sends $A_\mathcal{F}^0$ to $A_\mathcal{F}^1$ bijectively. Hence if $A_\mathcal{F}^0$ is meager then so does $A_\mathcal{F}^1$, so $2^\omega$ is a union of two meager sets, a contradiction.
There were misunderstanding on the question, and the OP asks the set $$A_1=\{f\in S \mid f^{-1}[2\mathbb{N}+1]\in\mathcal{F}\}$$ is meager or not (where $S$ is the set of all increasing functions from $\omega$ to $\omega$. I assume that the topology is given as the subspace topology of $\omega^\omega$.)
The answer is still negative by the similar argument: observe that $f^{-1}[2\mathbb{N}+1]\cup f^{-1}[2\mathbb{N}]=\mathbb{N}$, so one of $f^{-1}[2\mathbb{N}+1]$ or $f^{-1}[2\mathbb{N}]$ must be a member of $\mathcal{F}$.
Now assume that the set $A_1$ is meager. We can see that the map $\phi(f)= f+1$ is an embedding from $S$ to itself. Moreover, $\phi[A_1]\supseteq\{f\in S\mid f(0)\ge 1\land f^{-1}[2\mathbb{N}]\in\mathcal{F}\}$. (This follows from $(g+1)^{-1}[2\mathbb{N}+1]=g^{-1}[2\mathbb{N}]$, and $f=g+1$ if $f(0)\ge 1$ for increasing $f$.)
Therefore, the set $\{f\in S \mid f(0)\ge 1\}$ is covered by two meager sets, namely $A_1$ and $\phi[A_1]$. However, this is impossible as $\{f\in S \mid f(0)\ge 1\}$ is a non-empty open set of $S$.