For a partial answer, let me prove that every strategically closed partial order admits a *nearly tactical* winning strategy, one that depends only on the previous two moves, that is, on the previous move of each player. (On first move, it uses only player I's initial move.)

**Theorem.** Every strategically closed partial order $\newcommand\P{\mathbb{P}}\P$ admits a nearly tactical winning strategy for player two, one that depends only on the two previous moves, one move for each player. 

**Proof.** Assume $\P$ is a strategically closed partial order, with winning strategy $\sigma$ for player II. 

For simplicity let me assume initially that every lower cone $p\downarrow=\{q\in\P\mid q\leq p\}$ in $\P$ has the same cardinality as $\P$. In this special case the lower cones are also equinumerous with the finite tuples from $\P$, and so we may attach labels to every condition $p$ with a finite sequence $p^{\to}=(p_0,\ldots,p_n)$, and furthermore, we may do this in such a way that every finite sequence occurs densely often as a label.

Let me now describe the nearly tactical strategy $\tau$. If the previous two moves were $p>q$, then we consider the label $p^{\to}$, and if this is a finite sequence that corresponds to having played the game as I am now describing, then we interpret $q$ as player I's move in that game, compute the response $r$ of $\sigma$ to that play, and then extend $r$ to a chosen condition $r^+$ whose label codes the sequence $p^{\to},q$. 

The point is that player II's moves will in effect code the playing history, and this is enough information to get by with nearly tactical strategy, since from the previous move of player II, we will be able to recover the entire play. That is, the next move of player I will be some $q^+$ below $r^+$, and player II will now be looking at $r^+,q^+$, from which he can compute $p^{\to},q$ and thus reconstruct the play according to $\sigma$. 

This nearly tactical strategy is therefore a winning strategy, because at the start of the game, we can ensure that the very first play by II starts the coding sequence properly to indicate that, and then at each subsequent play, it will build on that encoding, and so the tactical play will correspond to a play according to $\sigma$, and so it will have a lower bound. 

Note that although we had player II extend $\sigma$'s play from $r$ to $r^+$, in order to code the right play, nevertheless in the simulated $\sigma$ game, we can pretend player II played only $r$, and that player I had extended $r^+$ to the response $q^+$. That is, whenever one has a winning strategy, then it is also winning to play any stronger move than the winning strategy, since the extension can be absorbed into player I's next play.

Now finally let me explain how to handle the general case, without the homogeneous cardinality assumption on the lower cones of $\P$. Consider any strategically closed partial order $\P$. I claim that there is a dense suborder $\newcommand\Q{\mathbb{Q}}\Q$ in $\P$, such that every lower cone $q\downarrow$ in $\Q$ is at least as large as the corresponding upper cone $q\uparrow$. (We allow that these lower cones may have different cardinalities themselves.) To see this, consider first the dense suborder of conditions $p\in\P$ such that the lower cone $p\downarrow$ has a minimal cardinality — it cannot be made smaller by refining $p$. This is dense, since below any condition we may find one whose cone has smallest cardinality. Next, take a maximal antichain in that dense set, and let $\Q$ be the conditions in the dense set below a condition in that antichain. It follows that every downset in $\Q$ has size at least as large as the corresponding upset in $\Q$, since that upset is contained in the downset of the corresponding element of the antichain. 

Notice next that every dense suborder of a strategically closed order is also strategically closed, as player II can pretend that all moves occur in the suborder, simply by strengthening all moves of either player so as to be in the dense set, but otherwise following the original strategy. So $\Q$ is strategically closed. But now, since every lower cone $p\downarrow$ in $\Q$ is at least as large as the up set $q\uparrow$, we can label the nodes below $q$ with finite sequences from the upset, such that every descending play to $q$ occurs densely often below $q$. This is enough to implement the nearly tactical strategy above. And so every strategically closed partial order admits a nearly tactical winning strategy. $\Box$

This answer amounts to the partial-order analogue of the Debs result I mentioned in my other answer, which he proved for the topological Banach-Mazur games. I am unsure whether his argument is like mine or not, since I haven't yet been able to read his paper.