This is a linear algebra problem.
First note that there is no point in playing the same square twice. Each square should be played once, or not at all.
Consider the grid:
\begin{array}{|c|c|c|c|} \hline
a_{11}&a_{12}&a_{13}&a_{14}\\ \hline
a_{21}&a_{22}&a_{23}&a_{24}\\ \hline
a_{31}&a_{32}&a_{33}&a_{34}\\ \hline
a_{31}&a_{42}&a_{43}&a_{44}\\ \hline
\end{array}
Define $a$ as when $a_{yx} = 1$ you play the square, when $a_{yx} = 0$ you don't play the square. Then you can see that $a$ is the variable you want to solve for, and it is constrained by a set of linear equations shown below.
Let $b_{yx}$ be the initial values of the grid. Write out the statement "location y x becomes zero":
$b_{y, x} + \sum_{u,v}^{N,N} a_{u,v} \cdot (|u - x| + |v - y| \le_{1/0} D) \equiv 0 \pmod 2$
(Here I'm using $m \le_{1/0} n$ to mean "1 if $m \le n$ and 0 otherwise".)
...for example the the equation "the top left corner becomes zero for D=2" is:
$\begin{align}
b_{11}
& + a_{11}\cdot 1 + a_{12}\cdot 1 + a_{13}\cdot 1 + a_{14}\cdot 0 \\
& + a_{21}\cdot 1 + a_{22}\cdot 1 + a_{23}\cdot 0 + a_{24}\cdot 0 \\
& + a_{31}\cdot 1 + a_{32}\cdot 0 + a_{33}\cdot 0 + a_{34}\cdot 0 \\
& + a_{41}\cdot 0 + a_{42}\cdot 0 + a_{43}\cdot 0 + a_{44}\cdot 0 \equiv 0 \pmod 2
\end{align}$
This is a linear equation in $a$, the variable you need to solve for.
Build your $N^2 \times N^2$ matrix and solve for $a$:
$\begin {bmatrix} & \cdots \\ \vdots & \end{bmatrix}
\begin{bmatrix} a_{11} \\ a_{12} \\ \vdots \end{bmatrix} \equiv
\begin{bmatrix} -b_{11} \\ -b_{12} \\ \vdots \end{bmatrix} \pmod 2 $
Example as requested:
Consider the following initial position:
$$b = \begin{array}{|c|c|c|c|} \hline
1&0&0\\ \hline
0&1&0\\ \hline
0&0&1\\ \hline
\end{array}$$
Let $D = 1$.
The top left square is only affected by changes to 3 locations, $a_{11}, a_{12}, a_{21}$ :
$$1 + a_{11} + a_{12} + a_{21} \equiv 0 \pmod 2$$
The center square is affect by changes to 5 locations:
$$1 + a_{12} + a_{21} + a_{22} + a_{23} + a_{32}\equiv 0 \pmod 2$$
The bottom middle square is affected by changes to 4 locations:
$$0 + a_{22} + a_{31} + a_{32} + a_{33} \equiv 0 \pmod 2$$
And so on, you can get 9 equations total like this.
Write them as a matrix:
$$\begin{bmatrix}1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0\cr 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0\cr 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0\cr 0 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 0\cr 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1\cr 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0\cr 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1\cr 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1\end{bmatrix}
\begin{bmatrix} a_{11} \\ a_{12} \\ a_{13} \\ a_{21} \\ a_{22} \\ a_{23} \\ a_{31} \\ a_{32} \\ a_{33} \\\end{bmatrix} \equiv
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \pmod 2$$
Begin Reduced Row Echelon Reduction:
$$\begin{bmatrix}1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1\cr 1 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0\cr 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0\cr 0 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 0 & 1\cr 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0\cr 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 0\cr 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1\end{bmatrix}$$
$$\vdots$$
$$\begin{bmatrix}1 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1\cr 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & 1 & 0\cr 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\end{bmatrix}$$
$$\vdots$$
$$\begin{bmatrix}1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\cr 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0\cr 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\end{bmatrix}$$
Which corresponds to $a_{11} = 1$, $a_{22} = 1$, $a_{33} = 1$, and the other $a$ values are zero. And if you check, flipping those 3 will solve the problem.
A more general example. Suppose $N=3$ and $D=2$. The resulting matrix:
$$\begin{bmatrix}
1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & b_{11}\cr
1 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 0 & b_{12}\cr
1 & 1 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & b_{13}\cr
1 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 0 & b_{21}\cr
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & b_{22}\cr
0 & 1 & 1 & 1 & 1 & 1 & 0 & 1 & 1 & b_{23}\cr
1 & 0 & 0 & 1 & 1 & 0 & 1 & 1 & 1 & b_{31}\cr
0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & b_{32}\cr
0 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & b_{33}\end{bmatrix}$$
Which has reduced row echelon form:
$$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & b_{23}+b_{12} \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & b_{31}+b_{22}+b_{21}+b_{11} \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & b_{22}+b_{21} \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & b_{13}+b_{12} \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & b_{31}+b_{23}+b_{22}+b_{13}+b_{12} \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 & b_{22}+b_{11} \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & b_{22}+b_{12} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & b_{32}+b_{23}+b_{21}+b_{12} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & b_{33}+b_{31}+b_{23}+b_{21}+b_{13}+b_{11}
\end{bmatrix}
$$
Meaning that you only have a solution if
$$b_{32}+b_{23}+b_{21}+b_{12} \equiv 0 \pmod 2$$
$$b_{33}+b_{31}+b_{23}+b_{21}+b_{13}+b_{11} \equiv 0 \pmod 2$$
For the second example in your question, this is given by :
$$0+0+1+0 \equiv 0 \pmod 2$$
$$0+0+0+1+1+1 \equiv 0 \pmod 2$$
showing that your example is actually impossible.
