Define making a row "different" as coloring in the remaining square in the row with the opposite color of the square already colored in that row and making a row "similar" as coloring in the remaining square in the row with the same color of the square already colored in that row. WLOG let Player 1 be the person that goes first, and Player 2 be the other person. I claim that Player 2 will win if he: 1. makes all odd-numbered rows different 2. makes all even-numbered rows similar Consider the grid where all odd-numbered rows are different and all even-numbered rows are similar. Player 2 would win in this case and he can easily force Player 1 into this position because whenever Player 1 colors a square on a empty row, he can color the other square in the row and thus has the unique ability of making rows either different or similar. $\blacksquare$

We can see that the game will become a tie for $n=2$ quite easily. Now suppose $n\geq 3$. [asy][asy] for (int i=0 ; i<=2 ; ++i) {draw((0,i)--(3,i), grey);} for (int i=0 ; i<=3 ; ++i) {draw((i,0)--(i,2),grey);} pen border=black+2; draw(box((0,0),(1,2)),border); draw(box((1,0),(3,1)),border); draw(box((1,1),(3,2)),border); for (int i=0 ; i<=2 ; ++i) {draw((4,i)--(5,i),grey);} for (int i=4 ; i<=5 ; ++i) {draw((i,0)--(i,2),grey);} draw(box((4,0),(5,2)),border); [/asy][/asy] Denote the upper two shapes as $X$ and $Y$. Let $k=\left[\frac{n}{3}\right]$. Divide the $2\times n$ grid into $k$ of the $X$ shapes and $n-3k$ of the $Y$ shapes. So the $2\times n$ grid is now divided into $n$ dominos, and $B$ can play so that every domino has two cells of different color. Let's count the number of pairs of neighboring cells of different color. There are $n$ such pairs inside the dominos. Now denote the set of points that are included in the edges of both vertical dominos and horizontal dominos as $V$. For every point $v$ in $V$, there is an additional pair of neighboring cells of different color, other than that inside of the vertical domino. Since $\left|V\right|=2k-1$ for $n=3k$, and $\left|V\right|=2k$ otherwise, the number of pairs of neighboring cells of different colors are at least $5k-1$,$5k+1$,$5k+2$ for $n=3k$, $n=3k+1$, $n=3k+2$ respectively. Recall that the total number of pairs of neighboring cells are $3n-2$. So we can now see that more than half of the pairs of neighboring cells are pairs of neighboring cells of different color, and so $B$ wins.