Is there an example of a poset $P$ that is a regular suborder of $Q$ such that $Q$ is $\omega_2$-strategically closed, but the quotient forcing $Q/P$ fails to be $\omega_1$-strategically closed?
To clarify, I am using the following definition, equivalent to the one in James Cummings' Handbook of Set Theory article:
For a partial order $\mathbb{P}$ and an ordinal $\alpha$, we define a game $G_\alpha(\mathbb{P})$ with two players Even and Odd. Even starts by playing some element $p_0 \in \mathbb{P}$. At successor stages $\beta+1$, the next player must play some element $p_{\beta+1} \leq p_\beta$. Even plays at limit stages $\beta$ if possible, by playing a $p_\beta$ that is $\leq p_\gamma$ for all $\gamma < \beta$. If Even cannot play at some stage below $\alpha$, the game is over and Odd wins; otherwise Even wins. We say that $\mathbb{P}$ is $\alpha$-strategically closed if for every $p \in \mathbb{P}$, Even has a winning strategy with first move $p$. Note that under this definition, every partial order is trivially $\omega$-strategically closed.
One example is the (square * thread) forcing. If $\kappa$ is regular, then there is a two-step iteration $P*Q$ such that $P$ forces $\square_\kappa$ in a $(\kappa+1)$-strategically closed way, and then $Q$ forces a continuation of the square sequence to $\kappa^+$, collapsing $\kappa^+$. The iteration has a $\kappa$-closed dense set, but $Q$ cannot be even countably strategically closed over $V^P$. The reason is that if it were, we could produce a complete binary tree following the strategy that decides initial segments of the thread $T$, such that any branch through the tree decides the thread up to some fixed ordinal $\alpha$, by recursively stretching out the decisions as needed. This gives continuum-many distinct possibilities for $T \cap \alpha$, but there’s only one: $C_\alpha$.