Yes, it has been studied by Johan Wästlund in A solution of two-person single-suit whist, which gives an efficient algorithm to compute the value of a position in this game (Theorem 10.1). He has also studied the more general situation of multiple suits, but with the restriction that each suit is split evenly between the two players, in Two-person symmetric whist. Here some familiar values from combinatorial game theory appear, reflecting the fact that breaking a new suit is generally disadvantageous to the player doing so.
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.
Best Answer
Update. I made a blog post about Infinite Sudoku and the Sudoku game, following up on ideas in this post and the comments below.
I claim that the second player wins the even-sized empty Sudoku boards and the first player wins for odd-sized empty Sudoku boards, including the main $9\times 9$ case. (The odd-case solution uses a key idea of user orlp in the comments.)
Consider first the even-sized board case, which is a little easier. For example, perhaps we have a board of size $16\times 16$, divided into subsquares of size $4\times 4$.
The second player can win in this case by the mirror-play strategy. (This argument was pointed out to me by my daughter, 11 years old.) That is, given any move by the other player, let the second player play the mirror image of that move through the origin. The new play cannot violate the Sudoku condition if the previous play did not, since the new violation would reflect to an earlier violation. The point is that on an even-sized board, the reflection of any row, column or subsquare will be a totally different row, column or subsquare, and so by maintaining symmetry, the second player can ensure that any violation of the Sudoku conditions will arise with the first player.
This copying strategy breaks down on the odd-sized board, however, including the main $9\times 9$ case, since there is a central row and column and a central subsquare and copying a move there would immediately violate the Sudoku conditions.
Nevertheless, user orlp explained in the comments how to adapt the mirroring strategy to the odd case.
Namely, in the main $9\times 9$ case, let's have the first player play a $5$ in the center square, and thereafter play the ten's complement mirror image of the opponent's moves. That is, if opponent played $x$, the first player should play $10-x$ in the mirror location. In this way, the first player can ensure that after her moves, the board is ten's complement symmetric through the origin. This implies that any violation of the Sudoku requirement will reduce by reflection to an earlier violation in the reflected moves, and so it is a winning strategy.
More generally, in the general odd case $k^2\times k^2$ for $k^2=2n-1$, player one will play $n$ in the middle square, and then proceed to play the $2n$'s complement mirroring move of the opponent. In this way, the first player ensures that after her play, the board remains $2n$'s complement symmetric, and this implies that she will not be the first to violate the Sudoku conditions. So it is a winning strategy.
Notice that in the even case, the second player could also have won by playing the complement mirror strategy, rather than the mirror strategy, since again any violation of the Sudoku condition would reflect to an earlier but complementary violation.
Finally, see my blog post for the winning strategy in the case of the Infinite Sudoku game, which came up in the comments.