I don't know about the game attributed to Specker, but here is a simple game with your desired features.
Let us call it the Chocolatier's game. There are two players, the Chocolatier and the Glutton. To begin play, the Chocolatier serves up finitely many unique and exquisite chocolate creations on a platter, and then the Glutton chooses one of them to eat. Play continues — at each stage the Chocolatier adds finitely many additional chocolates to the platter, and the Glutton consumes one of them. The chocolates accumulate on the platter as play progresses (except for those that have been eaten).
After infinite play, the Glutton wins if every single chocolate that was served was eventually consumed. Otherwise, the Chocolatier wins.
The Glutton can easily win simply by paying attention to the order in which the chocolates were added, and consuming them in that order. If only finitely many chocolates are added at each stage, he should simply make sure to consume them before moving on to the chocolates that were added at later stages.
Indeed, the Glutton can win even if the Chocolatier places countably many chocolates on each turn. At turn $n=2^k(2m+1)$, let the Glutton eat the $k$th chocolate added at stage $m$, if any, and otherwise eat arbitrarily.
But what about strategies that depend only on the current position, that is, the assortment of chocolates on the platter?
In one sense, it is easy to see that there can be no such winning strategy for the Glutton. If the Glutton will choose a particular chocolate from a given assortment, then let the Chocolatier simply replace it with an identical chocolate type on the next move, and again on all subsequent moves. If the strategy does not know the history, then the Glutton will choose it again every time, and the other chocolates will never be eaten.
It makes a more interesting game to say that the Chocolatier loses if chocolate types are ever repeated. And for this version of the game, there are some interesting things to say.
If there are only countably many chocolate types available in all, then again the Glutton has a winning strategy that depends only on the current assortment on offer. Namely, the Glutton should fix an enumeration of the possible chocolate types that might appear, and at each stage select the chocolate that appears earliest in this order — it is the tastiest-looking chocolate as defined by that priority. With this strategy he will succeed in eating all the chocolates, because at the limit, if any chocolate was left, there would have to be a tastiest-looking one (earliest in the enumeration), and this would have been eaten once the finitely many tastier chocolates had been consumed.
If the Chocolatier were uncountably creative, however, and able to serve up uncountably many different chocolate types, then this argument breaks down. Indeed, in this case I claim there is no winning strategy for the Glutton that depends only on the chocolate assortment on offer. And indeed, there is no such strategy that works even if we should insist that the Chocolatier present an assortment only of size two at each stage.
To see this, suppose the Glutton will follow a fixed strategy that selects a particular chocolate to eat from amongst any two chocolates.
For any given chocolate $c$, if there are infinitely many others $d_n$ that would be preferred to it by the strategy, if presented as a pair $\{c,d_n\}$, then the Chocolatier can present these pairs $\{c,d_0\}$, $\{c,d_1\}$, $\{c,d_2\}$, in turn. At each stage, $d_n$ would be consumed and the Chocolatier can present $\{c,d_{n+1}\}$. In the end, the inferior chocolate $c$ would never have been selected and so the Chocolatier will win.
So if this is a winning strategy for the Glutton, then we may assume that every chocolate is preferred to all but finitely many of the others. But this is simply impossible with an uncountable set. To see this, take any countably infinite set of chocolates and close under the finite witnesses of strictly preferred chocolates. One thereby constructs a countably infinite set $D$ of chocolates so that any chocolate in $D$ is preferred to any chocolate not in $D$. Since there were uncountably many chocolate types, there is some $c\notin D$. This contradicts our assumption that there are only finitely many chocolates preferred to a given chocolate.
Let me remark that in the version of the game where we do not allow the Chocolatier to repeat chocolate types (as opposed to saying that this is allowed, but cause him to lose), then we should really include the list of already-consumed chocolates as part of the position, and this complicates the arguments above. I suspect but do not yet know that the Glutton can have no winning strategy that depends only these more general kinds of positions in the uncountable case.