Let $A$ be a non-singular $n$-by-$n$ matrix with integer entries. Assume that $p^r\nmid \det(A)$. Does it follow that $A$ has an $(n-r+1)$-by-$(n-r+1)$ minor that is non-singular modulo $p$?
If the answer is no: what if we put some additional conditions -- say, $p$ large compared to $n$, and/or $A$ having bounded entries?
Let $s = \operatorname{ord}_p(\det(A))$, i.e. $p^s\ \| \det(A)$. Note that $|\det(A)| = \# \operatorname{coker}(A \colon \mathbf Z^n \to \mathbf Z^n)$. Right exactness of the tensor product shows that $\operatorname{coker}(A \otimes \mathbf F_p) \cong \operatorname{coker}(A) \otimes \mathbf F_p$, so $\operatorname{coker}(A \otimes \mathbf F_p)$ has cardinality at most $p^s$, hence dimension at most $s$. This should answer your question (affirmatively).
I think this is all about Smith Normal Form. Write $XAY = {\rm diag}(d_{1},d_{2},\ldots, d_{n})$ where $X$ and $Y$ are unimodular and where the $d_{i}$ are integers such that $d_{i} | d_{i+1}$ for each $i$. There are various terminologies for the $d_{i}$ sometimes conflicting. Let me call them the determinantal divisors (sometimes they are called elementary divisors sometimes invariant factors). In any case, they are unique up to sign and they determine and are determined by the structure of the Abelian group $\mathbb{Z}^{n}/{\rm Im}A$ when $A$ acts by multiplication on $\mathbb{Z}^{n}$ identified with $n$ long integer column vectors.
Your hypotheses imply that $p \not | d_{i}$ for $1 \leq i \leq n-(r-1).$ On the other hand, it is "well-known" that for $1 \leq i \leq m$, the product $d_{1}d_{2} \ldots d_{m}$ is the gcd of all the $m \times m$ minors of $A$. Hence your assumptions imply that there is indeed some $(n-r +1) \times (n-r+1)$ minor of $A$ which is not divisible by $p$.
