Assume the contrary, that every $2 \times 2$ square contains an even number of black cells. Let us compare two adjacent rows.
If the first cell in the lower row is the same color as the first cell in the upper row (e.g. black), then the second cells in these rows are also the same color (either black or white). In the same way we conclude further and see that these two rows are colored exactly the same.
If the first cell in the lower row is the opposite color from the first cells in the upper row, then the second cells in the lower row is also the opposite color from the second cells in the upper row. The same reasoning shows that the color of each cell in the lower row is the opposite color from the corresponding cell in the upper row.
Hence all the rows that begin with a black cell are equal, as are all the rows that begin with a white cell.
Let $a$ be the number of rows that begin with a black cell. Then $1000 - a$ rows begin with a white cell. Let $d$ be the difference between the number of black and the number of white cells in rows that begin with a black cell. Then the difference between the number of black and the number of white cells in rows that begin with a white cell is $-d$.
The difference between the total number of black and the total number of white cells on the board is
$$a \cdot d + (1000 - a) \cdot (-d) = 2ad - 1000d = (2a - 1000)d.$$
Hence we have $(2a - 1000)d = 2012$.
Notice that $d$ is even, since the total number of black and white cells in each row is $1000$. Also, it is clear that $d \le 1000$. Therefore from $(a - 500)d = 2 \cdot 503$ follows $d = 2$ so $a = 1003$ which is clearly not possible.