$GL_n(\Bbb Z_p)$ conjugacy classes in a $GL_n(\Bbb Q_p)$ conjugacy class
Is it always finite? No.
The matrices $u_n=\begin{pmatrix}1 & p^n\\0 & 1\end{pmatrix}\in\mathrm{GL}_2(\mathbf{Z}_p)$ are all conjugate in $\mathrm{GL}_2(\mathbf{Q}_p)$, and pairwise non-conjugate in $\mathrm{GL}_2(\mathbf{Z}_p)$.
Indeed, a matrix in $\mathrm{GL}_2(\mathbf{Q}_p)$ conjugating $u_n$ to $u_{n+m}$, for $m\ge 1$, is necessarily of the form $\begin{pmatrix}p^m\lambda & p^m\lambda\mu\\0 & \lambda\end{pmatrix}$ for scalars $\lambda\neq 0$, $\mu$. If it is in $\mathrm{GL}_2(\mathbf{Z}_p)$, the determinant $p^m\lambda^2$ has to be of valuation $0$, hence $\lambda$ has valuation $-m/2<0$ and $p^m\lambda$ has valuation $m/2>0$, and hence the trace $\lambda+p^m\lambda$ also has valuation $-m/2$, contradiction since the trace should be in $\mathbf{Z}_p$.
I'd be more optimistic for semisimple conjugacy classes.
Since François proved the finiteness for semisimple conjugacy classes, let me complete the proof of the converse: namely that if the $\mathrm{GL}_d(\mathbf{Q}_p)$-conjugacy class of $A\in M_d(\mathbf{Q}_p)$ splits into finitely many $\mathrm{GL}_d(\mathbf{Z}_p)$-conjugacy classes, then $A$ is semisimple.
Indeed, first if $P$ is a nonzero polynomial, the intersection of $\mathrm{Ker}(P(A))$ with $\mathbf{Z}_p^d$ is a direct summand of the latter as $\mathbf{Z}_p$-module. Hence, up to conjugation by some element of $\mathrm{GL}_d(\mathbf{Z}_p)$, we can suppose that $A$ is block-upper triangular $$A=\begin{pmatrix} A_1 & \dots &\\ 0 & \ddots & \vdots \\ 0 & 0 &A_k\end{pmatrix}$$ such that for each diagonal block $A_i$ there is some irreducible polynomial $P_i$ over $\mathbf{Q}_p$ such that $P_i(A_i)$. Let $A'$ be the block-diagonal matrix with the same diagonal blocks $A_i$, so $A'$ is semisimple. Let $u$ be the diagonal matrix that equals $p^{k-i}$ on the $i$-th block. Then $\lim_{n\to\infty}u^iAu^{-i}=A'$.
Assume the finiteness property. Since the $\mathrm{GL}_d(\mathbf{Z}_p)$-conjugacy classes are compact and there are finitely many in the $\mathrm{GL}_d(\mathbf{Q}_p)$-conjugacy class $\mathcal{C}$ of $A$, we deduce that they are open in $\mathcal{C}$. In particular, for $n$ large enough, $u^iAu^{-i}$ belongs to the $\mathrm{GL}_d(\mathbf{Z}_p)$-conjugacy class of $A'$. This proves that $A$ is semisimple.
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