A non-coloring solution:
No, it's impossible (every $v_n$ variable is the number of the vertical blocks starting on row $n$):
11 can be written as $4k+3$ and 10 as $4k+2$, with each horizontal block taking up 4 squares on a row. The rest are vertical blocks.
Row 1: $v_1 ≡ 3$ $(mod$ $4)$
Row 2: $v_1 + v_2 ≡ 3$ $(mod$ $4)$
Row 3: $v_1 + v_2 + v_3 ≡ 3$ $(mod$ $4)$
Row 4: $v_1 + v_2 + v_3 + v_4 ≡ 3$ $(mod$ $4)$
Row 5: $v_2 + v_3 + v_4 + v_5 ≡ 3$ $(mod$ $4)$
Row 6: $v_3 + v_4 + v_5 + v_6 ≡ 2$ $(mod$ $4)$
Row 7: $v_4 + v_5 + v_6 + v_7 ≡ 3$ $(mod$ $4)$
Row 8: $v_5 + v_6 + v_7 + v_8 ≡ 3$ $(mod$ $4)$
Row 9: $v_6 + v_7 + v_8 ≡ 3$ $(mod$ $4)$
Row 10: $v_7 + v_8 ≡ 3$ $(mod$ $4)$
Row 11: $v_8 ≡ 3$ $(mod$ $4)$
This leads to the contradiction where $v_3+v_4+v_5+v_6$ should have been 2 modulo 4 based on the middle row, but is actually divisible by 4.
Taking away a square:
Let the single square be below the middle row. Then $v_1$ and $v_5$ can be written in the $4k+3$ form. $v_2, v_3, v_4$ and $v_6$ can be written in the $4k$ form. Rows 8 and 9 (from top to bottom) give different results, so it's on Row 8 because its mod 4 sum is 1 lower. Same goes for the column. Row 8, column 8, rotations are welcome.