Assume AC. Suppose $X$ is a subset of the irrationals (Baire Space) for which neither player has a winning strategy (i.e. the game $G(\omega, X)$ is not determined). Is $X$ nonmeasurable in the Lebesgue sense as a subset of $\mathbb{R}$?
(My argument is somewhat easier if you consider games where the players play $0$s and $1$s, so that the payoff set is in Cantor space $2^\omega$, and we use the usual coinflipping probability measure; but an essentially similar idea works in Baire space.) For any game with payoff set $A$, where player I wins if the play is in $A$, consider the following slightly modified game $A^\ast$, which is just like $A$, except we insert a pair of dummy moves between each pair of actual moves, and insist that player I play a $0$ in this dummy round, while player II can play anything. Thus, a sequence or play is in the payoff set $A^\ast$ if indeed that sequence shows that player I did play a $0$ in all the dummy rounds (so every fourth digit is $0$), and furthermore, if we omit the dummy rounds entirely from the sequence, we get a sequence in $A$. Thus, playing the game $A^*$ is just like playing $A$, except that the play is interrupted for these silly dummy rounds. Note that player I has no incentive not to play a $0$ on those rounds, and player II's plays in the dummy rounds are ignored entirely. Thus, it is clear that a player has a winning strategy for $A$ if and only if he or she has a winning strategy for $A^\ast$, since we can translate the strategies from $A$ to $A^\ast$ and back again. The dummy rounds really don't change the difficulty of winning the game. But the point now is that because every fourth digit of $A^\ast$ is $0$, it follows that $A^\ast$ has measure $0$. (Every time you insist that a particular digit is $0$, it cuts the measure in half again.) The conclusion, therefore, which does not use the axiom of choice, is that if there is a nondetermined set, then there is a nondetermined set with measure $0$. In particular, there is a nondetermined set that is measurable. 


I have two possible answers. The first is short and possibly not the one you want. The second is probably the right one. First: take any (co)analytic subset $X$ of the Baire space $\omega^{\omega}$, which is not Borel. Than it is consistent with ZFC that the game $G(\omega,X)$ is not determined (ZFC just proves the determinacy of Borel games), but $X$ is Lebesguemeasurable (in fact universally measurable). I guess this is a consistent proof of "no", to your question. Second Assume AC. Then there exists a universallynull set (hence Lebesgue null, since the Lebesgue measure is atomless) $X$ of cardinality $\geq\aleph_{1}$. Then one can show that such a set $X$ can not be a Perfect set. Now let us consider the socalled Perfectset game $PSG(\omega, X)$, which is technically a Gale Stewart game $G(\omega, Y)$, with $Y$ of about the same complexity of $X$ (in particular $Y$ is universallynull nonperfect set of cardinality $\geq\aleph_{1}$). This game is not determined since $Y$ is not Perfect nor countable by construction, and it is knonw that such a game is determined only if $Y$ is Perfect or countable. Yet $Y$ is universally null, hence Lebesgue measurable. This second example is mentioned in Martin's "Blackwell's determinacy", where it is credited to Greg Hjorth. This is a proof in ZFC of "no", to your quesiton. 

