I guess the answer is that
with best play on both sides the scores are equal (for any value of 7)
but I don't have anything resembling a proof. Here are some very partial results.
For a $1\times n$ board,
the scores (player 0 minus player 1) go 0,2,2,0,0,6,6,0,0,10,10,0,0,14,14, etc. (This is pure calculation; all games are equivalent.)
For a $2\times n$ board,
the scores go 2,0,2,0,2,4,6, ... (by computer calculation) and I'm not sure what the pattern is. These scores are always non-negative, because player 0 has the following strategy that guarantees getting both rows: play in the less-full row until at least one row has only one space left; if that row has an odd number of 1s, play in that row; otherwise, play only in the other row from then on (which player 0 can always do because when it's her turn the number of spaces in the other row is odd and hence nonzero).
For other small board sizes,
3x3 and 3x4 are both draws (by computer calculation) and I haven't gone any further than that.
When either dimension of the board is a multiple of 4,
player One can be sure of at least drawing: suppose e.g. the horizontal dimension is a multiple of 4, then when player 0 plays at (x,y) player 1 plays at (x XOR 1, y) ensuring that every row contains exactly as many 0 as 1, hence an even number of both, so all the rows score for player 1, so player 1 at least draws.
I have not found any case
where One scores more than Zero (i.e., where the scores I've been reporting above are negative)
and suspect there are no such cases but have no proof. (I briefly thought I had a strategy that lets player 0 ensure that all rows have odd sum, which would certainly do it, but obviously there cannot be such a strategy unless the total number of 1s is odd, which e.g. it is not on any square board.)