If we consider finite deterministic games, then there is always an algorithm that calculates if one of the two players have a winning strategy or one of the two players can force a draw (just make a full search on the game space) ... the problem (as noted by jmite) is that it requires too much time and space :-)
But if we consider games that are "unlimited" (in some way), then there are two-person win-lose games (rather artificial) with decidable rules but no computable winning strategies.
A classical example is the following: given a computable function $f: \mathbb{N}^{m} \to \{A,B\}$, the two players alternately choose a natural number $x_i$ (player A begins). The games ends after $m$ turns (suppose that $x_1, x_2, ..., x_{m}$ are the picked numbers); the winner is player $A$ if and only if $f(x_1, x_2, ..., x_{m}) = A$. For this game, given $m$ and $f$, there is no algorithm that can tell you if A (or B) has a winning strategy. Rabin [Rab57] proposed a similar simple three moves game (m = 3) for which player B has a winning strategy, but it is not computable.
A still open problem (as far as I know it is still open :-) is a chess game played on an infinite board:
Given a finite set of chess pieces and their position on an infinite bidimensional $\mathbb{Z} \times \mathbb{Z}$ board, is there a winning strategy for white?
For other references see: J. P. Jones, Some undecidable determined game
(I'm not an expert, and I would like to know if there are more natural undecidable games)
[Rab57] Michael O. Rabin, Effective computability of winning strategies, Contributions to the theory of games, vol. 3, Annals of Mathematics Studies, no. 39, Princeton University Press, Princeton, N. J., 1957, pp. 147-157.