Let's say $m$ is the number of rows and $n$ is the number of columns.
If $m$ is odd, then we can find a simple valid coloring by coloring alternate rows black and white (except for the black borders of course). If $n$ is odd, we can do the same thing with the columns to find a valid coloring.
The case remains where $m$ and $n$ are both even. Claim: there is no valid coloring in this case.
Proof: Assume the claim is false, i.e. there is a valid coloring when $m$ and $n$ are both even.
Consider the interior corners of the grid, and color them red and blue in a checkerboard pattern as in the image below. Now construct a graph using these corners as vertices, connecting two corners along a grid edge if and only if the edge separates a black square and a white square.
Now for each vertex of this graph, the colouring rules enforce that exactly two edges are incident to this vertex. Furthermore, every edge connects a red vertex with a blue vertex.
So the number of edges is exactly twice the number of red vertices (there are two edges for each red vertex), and the number of edges is also exactly twice the number of blue vertices. That means there are as many red vertices as blue vertices, so the total number of vertices is even.
But wait, that's impossible! Because the total number of vertices is exactly $(m-1)(n-1)$, which is a product of two odd numbers and therefore odd. We have a contradiction, so our last assumption is false, therefore the claim is true.
Alternate contradiction: Ignoring the vertex colors, if there are two vertices incident to each edge and two edges incident to each vertex, then the number of vertices and edges is the same. There's an odd number of vertices, so there's an odd number of edges.
However, in each column of squares, there's an even number of horizontal edges, since the squares change between black and white an even number of times in the row. Similarly, in each row of squares, there's an even number of vertical edges. Summing all these even numbers, we count each edge exactly once, so the total number of edges must be even. That's a contradiction.
Best Answer
For rectangles, if there exists an $a\times b$ rectangle with $a\neq b$ such that it contains at least four black squares, then its transpose $b\times a$ will certainly be a distinct rectangle with at least four black squares that will fit on the chessboard. Similarly, any rectangle such that its transpose is a distinct rectangle will have the property that reversing the colors within the rectangle is possible by shifting it (for example, the only two $7\times 8$ rectangles on the board are both distinct, as one has a white square in the upper left corner while the other has a black square in the upper left corner). Thus for any $a\times b$ rectangle that has at least four white squares, reversing the colors (by shifting one row up or down as needed) will contain at least four black squares. This is true for any $a\times b$ rectangle that possesses at least four black squares except $1\times 7$ rectangles. A $1\times 7$ rectangle that possesses four black squares only contains three white squares.
Next, let's consider squares. The smallest $a\times a$ square that contains at least four black squares is the $3\times 3$ square. This contains either 5 black and 4 white or 4 black and 5 white. Its transpose will be nondistinct. But, shifting it one row or column will reverse the colors and be distinct. Thus, each such square will have a second "copy" except for the $8\times 8$ square because it cannot be shifted any rows or columns to reverse the colors.
This yields:
$1\times 7, 7\times 1$: 2 distinct, count each once
$1\times 8, 2\times 4^+, 3\times 4^+, 4\times 5^+, 5\times 6^+, 6\times 7^+, 7\times 8$: 21, count each four times.
$3\times 3, 4\times 4, 5\times 5, 6\times 6,7\times 7$: 5, count each twice.
$8\times 8$: Count once.
For this count, I am using $2\times 4^+$ to represent all distinct pairs where the second number is at least 4, thus $2\times 4, 2\times 5, 2\times 6, 2\times 7, 2\times 8$ are all represented by this shorthand.
This yields $2+21\cdot 4+5\cdot 2+1 = 97$, which is the answer you arrived at.