Let $\mathbb{Z}_p$ be the ring of $p$-adic integers and $\overline{\mathbb{Q}}_p$ a fixed algebraic closure of the field $\mathbb{Q}_p$ of $p$-adic numbers. Let $R$ be a complete Noetherian local $\mathbb{Z}_p$-algebra. It is not hard to see that if $R$ is a finite $\mathbb{Z}_p$-algebra, then the set $\operatorname{Hom}_{\mathbb{Z}_p}(R,\overline{\mathbb{Q}}_p)$ of continuous homomorphisms of $\mathbb{Z}_p$-algebras is finite. A natural question, then, is the reverse also true?
Let $R$ be a complete Noetherian local $\mathbb{Z}_p$-algebra of characteristic zero such that the set $\text{Hom}_{\mathbb{Z}_p}(R,\overline{\mathbb{Q}}_p)$ is finite. Is it true that $R$ is a finite $\mathbb{Z}_p$-algebra?
(Added: this answer is implicitly assuming that $R$ has residue field $\mathbf{F}_p$. From context, I am guessing you are thinking about Galois deformation rings where this is always satisfied. Otherwise $\mathbf{Q}_p$ is also a counter example as Torsten Schoeneberg points out. The assumption on residue field implies that $R$ is a quotient of $\mathbf{Z}_p[[X_1,\ldots,X_d]]$ for some $d$ which is what is being implicitly used below.)
The weaker statement that $R[1/p]$ is finite over $\mathbf{Q}_p$ is true (so if $R$ is flat over $\mathbf{Z}_p$ then you are fine.)
Certainly we can assume that the Noetherian ring $R[1/p]$ has only finitely many maps to $K = \overline{\mathbf{Q}}_p$. The point is that $R[1/p]$ is a Jacobson ring (see e.g. http://math.stanford.edu/~conrad/modseminar/pdf/L04.pdf), and that any maximal ideal $P$ of $R[1/p]$ comes from a map $R[1/p] \rightarrow K$. So to show that $R[1/p]$ is finite, it suffices to show that it is Artinian. Since $R[1/p]$ is Noetherian, it suffices to show that any prime ideal is maximal. But assume that $\mathfrak{p}$ is prime. Then by the Jacobson property, $\mathfrak{p}$ is the intersection of the maximal ideals $P_i$ which contain it. But we are assuming there are only finitely many such maximal ideals, so
$$R[1/p]/\mathfrak{p} \simeq R[1/p]/\bigcap P_i \simeq \prod R[1/p]/P_i,$$
where the second isomorphism is the Chinese Remainder Theorem (which one can apply since the intersection is finite). Since the LHS is a domain, this clearly shows that $\mathfrak{p} = P_i$ for some $i$ and is thus maximal.
It's not true that $R$ itself has to be finite over $\mathbf{Z}_p$; if
$$R = \mathbf{Z}_p[[x]]/(px),$$
then any map from $R$ to $K$ sends $x=0$ so there is only one such point. But $R $ surjects onto $\mathbf{F}_p[[x]]$ so $R$ is not finite over $\mathbf{Z}_p$.