The answer is yes if $A \in \mathrm{GL}_n(\mathbb{Z}_p)$ is semisimple.

We may think of a matrix $A \in M_n(\mathbb{Z}_p)$ as a $\mathbb{Z}_p$-lattice of rank $n$ endowed with a $\mathbb{Z}_p$-linear endomorphism: take the standard lattice $L=\mathbb{Z}_p^n$ endowed with the endomorphism $\varphi$ associated to $A$. Clearly, two matrices $A,A' \in M_n(\mathbb{Z}_p)$ are $\mathrm{GL}_n(\mathbb{Z}_p)$-conjugated if and only if the associated pairs $(L,\varphi)$ and $(L',\varphi')$ are isomorphic.

More formally, consider the algebra $R = \mathbb{Z}_p[A]$ inside $M_n(\mathbb{Z}_p)$. Then $A$ gives rise to an $R$-module $M$ which is $\mathbb{Z}_p$-free of rank $n$. Note that if $A' \in M_n(\mathbb{Z}_p)$ is another matrix which is $\mathbb{Q}_p$-conjugated to $A$ then the algebras $\mathbb{Z}_p[A]$ and $\mathbb{Z}_p[A']$ are isomorphic, as they are both isomorphic to $R = \mathbb{Z}_p[X]/(\mu)$ where $\mu$ denotes the minimal polynomial. However, the associated $R$-modules $M$ and $M'$ need not be isomorphic. In fact, they are if and only if $A$ and $A'$ are $\mathbb{Z}_p$-conjugated.

So the problem reduces to classifying the $R$-modules which are $\mathbb{Z}_p$-free of rank $n$ up to isomorphism. In the semisimple case, the following theorem of Jordan-Zassenhaus gives a positive answer to your third question.

**Theorem** (Jordan-Zassenhaus). Let $K$ be a local or global field with ring of integers $\mathcal{O}_K$. Let $L$ be a (commutative) semisimple $K$-algebra, and let $R$ be an $\mathcal{O}_K$-order in $L$. Then for any integer $n \geq 1$, the set of isomorphism classes of $R$-modules which are $\mathcal{O}_K$-lattices of rank $\leq n$ is finite.

The case at hand follows by taking $K=\mathbb{Q}_p$, $L=\mathbb{Q}_p[X]/(\mu)$ and $R=\mathbb{Z}_p[X]/(\mu)$ where $\mu$ is a squarefree polynomial. A proof of the Jordan-Zassenhaus theorem is given in Reiner, *Maximal orders*, (26.4), p. 228. It is stated only for global fields, but you can check it also holds for local fields.

**EDIT.** The proof is not difficult: first one treats the case where $R$ is a maximal order, which is clear since it is a product of discrete valuation rings (hence PIDs). If $R' \subset R$ is an arbitrary order and $M'$ is an $R'$-lattice then there are only finitely many possibilities for $M=M' \otimes_{R'} R$, and since $p^N M \subset M' \subset M$ there are only finitely many possibilities for $M'$. So we get the finiteness.

Regarding the question of an explicit classification, you may be interested by Marseglia's article *Computing the ideal class monoid of an order*, where he gives algorithms to compute explicit representatives (in the global case).

**EDIT 2**. Another article giving an algorithm in the global case appeared on the arXiv today: https://arxiv.org/abs/1811.06190