First, we make some observations about tables containing no irreducible $2\times 2$ subtables. Note that a $2\times 2$ subtable is irreducible iff it contains an odd number of $-$'s. Claim: If we are given the elements of the table that lie on the topmost row and the leftmost column, we can construct a unique table containing only reducible $2\times 2$ squares. Proof: Given any three elements of a $2\times 2$ square, there is a unique way to fill in the fourth and cause the square to be reducible. Thus, we can just work our way across the columns and down the rows, filling in elements whenever they are part of a $2\times 2$ square that already contains three other filled elements. Claim: Given any combination of rows and columns to switch, there is exactly one equivalent combination that does not flip the leftmost column. Proof: If the given composition does flip the leftmost column, then flip every column and row in the grid. This doesn't change any elements, as they are all flipped twice, but now the combination of rows and columns does not include the leftmost column. To prove that there is only one such combination fitting the requirements, we consider the leftmost column. By looking at which elements in that column are flipped, we can deduce which rows must have been included in the combination; then, by looking at the top row, we can deduce which columns must have been included. Thus, each combination is determined exactly. Now, the denouement. The number of tables containing no $2\times 2$ irreducible subtables is the same as the number of ways to put elements in the topmost row and leftmost column; that is, there are $2^n^+{}^m{}^-{}^1$ such tables. The number of unique ways to choose a combination of rows and columns to flip is the same as the number of ways to flip any combination of rows and columns that does not include the leftmost column; that is, there are $2^n^+{}^m{}^-{}^1$ such ways. Thus, there is a bijection between legal combinations and tables containing only reducible $2\times 2$ subtables, so every such table is clearable, and we are done.

A $4\times 4$ subtable is reducible iff there are an even number of elements contained in the X-marked squares: OXXO XOOX XOOX OXXO Thus, given any 7 elements in a ring like the ring of X-marked squares above, we can uniquely choose the 8th. We can count the total number of tables containing only reducible $4\times 4$ subtables using the slick argument that the parities of the 8-element sets are linearly independent, so we lose exactly one factor of 2 for each one. Since there are (3-n)(3-m) such $4\times 4$ subtables, there are $2{}^3{}^m^+{}^3{}^n{}^-{}^9$ tables containing only reducible subtables. In fact, we could probably also use this sort of argument when counting the number of unique combinations of rows, columns, and diagonals- just find the rank of the matrix and all that. Ideally, if we want the problem statement to be true, then the rank will be $3m+3n-9$, so the corank will be $7$. (If you don't know linear algebra: that means that, to finish up, we just have to find 7 rows/columns/diagonals such that not being allowed to switch any of them doesn't change the possible combinations of elements that we can flip, but ensures that any combination of flipping we can achieve must be achieved uniquely. That is left as an exercise to the reader, because this is supposed to be an outline!)