As commented by Jyrki Lahtonen, the statement is true and it is immediately implied by the following relation between the minors of $A^{-1}$ and the minors of $A$.
Proposition: If $A$ is an invertible $n\times n$ matrix, and if $i_1,\dots,i_n$ and $j_1,\dots,j_n$ be two permutations of $1,\dots,n$, then the minor of $A$ corresponding to rows $i_1,\dots,i_k$ and columns $j_1,\dots,j_k$, denoted by $d$, and the minor of $A^{-1}$ corresponding to rows $j_{k+1},\dots,j_n$ and columns $i_{k+1},\dots,i_n$, denoted by $d'$,satisfy that
$$d=\pm d'\det A.$$
Proof: Let $e_1,\dots,e_n$ be a basis of $\mathbb{R}^n$, and let $f_i= A e_i$, $i=1,\dots,n$. Then on the one hand,
$$\omega:=Ae_{j_1}\wedge\cdots\wedge Ae_{j_k}\wedge e_{i_{k+1}}\wedge \cdots\wedge e_{i_n}=\pm d\cdot e_1\wedge\cdots\wedge e_n,$$
on the other hand,
$$\omega=f_{j_1}\wedge\cdots\wedge f_{j_k}\wedge A^{-1}f_{i_{k+1}}\wedge \cdots\wedge A^{-1}f_{i_n}=\pm d'\cdot f_1\wedge\cdots\wedge f_n.$$
Since
$$f_1\wedge\cdots\wedge f_n=\det A\cdot e_1\wedge\cdots\wedge e_n,$$
the conclusion follows.
Best Answer
Let $\hat{A}$ be the altered matrix. We can write it as $\hat{A} = D A \bar{D}$ where $D$ and $\bar{D}$ are diagonal matrices all diagonal entries $\pm 1$. $D_{ii}=-1$ if row $i$ of $A$ is being negated and $\bar{D}_{ii}=-1$ if column $i$ of $A$ is being negated.
Any square submatrix of $\hat{A}$ is the product of a diagonal submatrix of $D,$ the same square submatrix of $A,$ and a diagonal submatrix of $\bar{D}.$ The submatrices of $D$ and $\bar{D}$ have determinant $\pm 1,$ the submatrix of $A$ has determinant in $\lbrace -1, 0, 1\rbrace,$ and the determinant of a matrix product is the product of the determinants.
So the answer is yes.