As Benjamin Steinberg mentioned in the comments, his paper On the endomorphism monoid of a profinite semigroup describes the analogous question for $End(G)$ where $G$ is a profinite semigroup.
His argument also shows that if $G,H$ are profinite with $G$ finitely generated, then $Hom(G,H)$ is a profinite space. For completeness's sake, below I have translated Benjamin Steinberg's argument into our setting.
His argument uses a form of Ascoli's theorem for uniform spaces:
Theorem (Ascoli). Let $X,Y$ be compact Hausdorff spaces equipped with their unique uniform structures (which are compatible with their topology) and let $C(X,Y)$ denote the space of continuous maps from $X$ to $Y$ equipped with the compact-open topology. Then for a family $F\subset C(X,Y)$, $F$ is compact (when given the induced topology) if and only if $F$ is closed and uniformly equicontinuous.
Here, recall that a subset $F\subset C(X,Y)$ is uniformly equicontinuous if for any entourage $E\subset Y\times Y$, $\bigcap_{f\in F}(f\times f)^{-1}(E)$ (a subset of $X\times X$) is an entourage for $X$.
Step 1 - $Hom(G,H)$ is closed inside $C(G,H)$
Let $f : G\rightarrow H$ be continuous but not a homomorphism. Then for some $g,g'\in G$, $f(gg')\ne f(g)f(g')$. Choose disjoint open neighborhoods $U,V$ of $f(gg')$ and $f(g)f(g')$ respectively. By continuity of multiplication we can find open neighborhoods $W$ of $f(g)$ and $W'$ of $f(g')$ such that $W\cdot W'\subset V$. Then let $\Phi\subset C(G,H)$ be the subset consisting of continuous functions $\phi : G\rightarrow H$ w/ $\phi(gg')\in U$, $\phi(g)\in W$, $\phi(g')\in W'$ - in particular $\phi$ is not a homomorphism. On the other hand $\Phi$ is visibly open in $C(G,H)$ and contains $f$. Since this holds for every continuous non-homomorphism, $Hom(G,H)$ is closed.
Step 2 - $Hom(G,H)\subset C(G,H)$ is uniformly equicontinuous
For an open neighborhood $U\ni 1_G$, let $E_U := \{(x,y)\in G\times G\;|\; xy^{-1}\in U\}$. Note that $E_{U\cap U'} = E_U\cap E_{U'}$. Recall that the uniform structure on $G$ is given by: A subset $E\subset G\times G$ is an entourage if and only if $E\supset E_U$ for some open neighborhood $U\ni 1_G$ (and similarly for $H$). We wish to show that for any entourage $E\subset H\times H$,
$$\bigcap_{f\in Hom(G,H)}(f\times f)^{-1}(E)$$
is an entourage (equivalently, contains an entourage) for $G$. Since $1_H\in H$ admits a neighborhood basis consisting of open normal subgroups and the set of entourages form a filter, it suffices to check this for $E = E_N$ where $N\le H$ is an open normal subgroup. In this case it is easy to check that $(f\times f)^{-1}(E_N) = E_{f^{-1}(N)}$, so it suffices to show that $\bigcap_{f\in Hom(G,H)} E_{f^{-1}(N)}$ contains an entourage of $G$. Indeed, $[G:f^{-1}(N)]\le [H:N]$, and since $G$ is finitely generated there are only finitely many open subgroups of index $\le [H:N]$ (this is Proposition 2.5.1 in Ribes-Zalesskii). Let $M$ denote their intersection, so $M\le G$ is also open. Thus we have
$$\bigcap_{f\in Hom(G,H)} E_{f^{-1}(N)}\supset \bigcap_{U\le_o G, [G:U]\le[H:N]} E_U = E_M$$
as desired.
Step 3 - $Hom(G,H)$ is compact
Thus $Hom(G,H)\subset C(X,Y)$ is closed and uniformly equicontinuous, so by the Ascoli theorem we find that $Hom(G,H)$ is compact. Let $H^G := \prod_{g\in G}H$ with the product topology, so $H^G$ is a compact Hausdorff and totally disconnected space. Let $i : Hom(G,H)\rightarrow H^G$ be the injection $f\mapsto (f(g))_{g\in G}$, then $i$ is clearly continuous, but since the source is compact and the target is Hausdorff $i$ is a homeomorphism onto its image. In particular, $Hom(G,H)$ is compact, Hausdorff, and totally disconnected (i.e., profinite).