For an $n$-cycle, player 1 wins if $n$ is odd, else player 2 wins.
Since all vertices have degree 2, any vertex can have at most 1 token on it. Token-carrying vertices (of different colors) can be grouped into connected components. Call a component occupied by a player if it contains some vertex colored by the player, and that two components separated by a single vertex are adjacent. For example, the 6-cycle $(-*-1-2-*-1-*-)$ has 2 components, the first occupied by both players, the second occupied by player 1, and they are adjacent.
Lemma: If less than $n-2$ tokens have been played, it is player $i$'s turn to play, and player $i$ occupies a component, then she can ensure that she continues to occupy a component until her next turn.
To do so, she simply topples a component occupied by her. Toppling an occupied component results in two adjacent components occupied by her (you can see this by working out the result of toppling an isolated component). These may merge with adjacent neighbors of the original component, but this does not affect her occupancy. The opponent can in her turn topple at most one of these two components, leaving the other still occupied.
Neither player can lose before $n-2$ tokens have been played, if both follow the above strategy. On the $(n-2)$th play, any move results in a single $(n-1)$-vertex component. Suppose $n$ is even, so that it is player 1's turn, and player 2 on the previous turn created two distinct components occupied by her as above. Whether player 1 topples one of the components if she also occupies it, or joins the two by playing on an empty vertex, it leads to a single connected component occupied by player 2. Player 2 then wins by toppling the final component on the $n$th move. The same argument holds for Player 1 if $n$ is odd.
The strategy as such does not necessarily work for paths, because toppling a component adjacent to an end of the path does not create two separate components. My hunch is that for even $n$, starting from the middle then playing as above may still be a winning strategy for player 1, as it avoids the ends till the last turn. I don't know what the right strategy is for odd $n$, though experimentation leads me to believe player 2 should win.
I also can't see if there's any way to generalize this for general graphs, because it depends crucially on there being at most one token per vertex.
This is only a heuristics, but I would assume that the game either ends after a few steps, or the outcome is determined simply by the parity of $(X+1)(Y+1)$ minus the number of gaussian primes in the rectangle. This is certainly the case if at the end of the game all non-primes are taken. So the player for which this heuristics predicts a win should try to play in such a way that all points of the rectangle are midpoints of lines with endpoints of different colour. Since most gaussian integers are not prime one of the basic heuristics of additive combinatorics predicts that the only way to avoid this is to have some extremely large Fourier coefficient, that is, the board looks like having red and blue stripes. But by taking midpoints of red and blue points one of the player can prevent this from happening, provided he has a sufficient number of choices left. This means that either the game ends early, or it should continue until all non-primes are taken, in which case the winner is determined by the aforementioned parity.
Best Answer
To begin with, one should think about how the game naturally splits into components, and what it means for a game position to be the “sum” of its components.
As the game proceeds, the set of unoccupied sites can separate into components that do not interact with each other. In such a situation, a move consists in making a move in one of the components (this is like the classical theory), and winning in one component means winning the entire game (this differs from the so-called normal playing convention where the game ends when a player is unable to move).
In classical combinatorial game theory, games are classified into four outcome classes: Positive (Left wins no matter who starts), Negative (Right wins), Fuzzy (Player to move wins) and Zero (Player not to move wins). In a game like connect-four that can end in a draw, there are nine outcome classes, namely the functions from {Red to move, Blue to move} to {Red wins, Draw, Blue wins}.
Just as in the classical theory, the set of positions is an abelian monoid under addition (position means what the board looks like, without information about who is to move), and one may define two positions $X$ and $Y$ to be equivalent if for every $Z$, the games $X+Z$ and $Y+Z$ belong to the same outcome class.
For instance, all positions with an even number of unoccupied sites, and where none of the players can win, are equivalent. These might be called “zero-positions”, since they include the empty position (neutral element of the monoid).
The additive structure carries over to addition of equivalence classes, but unlike the classical case, it doesn’t become a group. The zero class is a neutral element, but not all positions have inverses. For instance, if $X$ is a position is such that the player to move (whether Red or Blue) can win in one move, then there can be no other position $X’$ so that $X+X’$ becomes zero, because no matter what we add to $X$, the resulting game will still be a win for the player to move.
On the other hand some positions clearly do have inverses. There is a class that contains all positions with an odd number of free sites, and where nobody can win. We might call this class $\star$, since it is similar to the class "star" in the classical theory. These positions are clearly not equivalent to the zero positions, since adding them will turn a “zugzwang” into a first player win. But we do have the equation $\star + \star = 0$.
There are several games, in particular misère games and card games, where the analysis of a corresponding monoid leads to a solution, see for instance my paper The strange algebra of combinatorial games and its references.
A couple of questions: Is the connect-four monoid infinite? What do the connect-two and connect-three monoids look like?
Finally, it seems worth taking a look at the Masters thesis of Victor Allis from 1988: A Knowledge-based Approach of Connect-Four.