This is possible (filling in some details in the argument from my comment).
We use choice to fix an isomorphism $\mathbb C \cong \overline{\mathbb Q_p}$. Using this, we embed $\mathrm{SO}(3) \subset \mathrm{SO}_3(\mathbb C) \cong \mathrm{SO}_3(\overline{\mathbb Q_p})$.
We let $\mathrm{SO}_3 (\overline{\mathbb Q_p})$ act on the set $\mathrm{SO}_3 (\overline{\mathbb Q_p})/\mathrm{SO}_3( \overline{\mathbb Z_p})$, where $\overline{\mathbb Z_p}$ consists of the elements of $\overline{\mathbb Q_p}$ that are integral over $\mathbb Z_p$.
This action is faithful since an element acts trivially if and only if all its conjugates lie in $\overline{\mathbb Q_p}$$\mathrm{SO}_3( \overline{\mathbb Z_p})$, and conjugating by an element of a split maximal torus, we can see that this forces the element to lie in that torus, hence to lie in every split torus, thus lie in the center, which is trivial.
(If, instead of $\mathrm{SO}(3)$, we had a group like $\mathrm{SU}(2)$ with center, we could mod out by the integral elements congruent to $1$ mod $p$)
The set $\mathrm{SO}_3 (\overline{\mathbb Q_p})/\mathrm{SO}_3( \overline{\mathbb Z_p})$ is countable. Indeed, every class in it is represented by some element in $\mathrm{SO}_3 (\overline{\mathbb Q_p})$, which must lie in $\mathrm{SO}_3(K_p)$ for some finite extension $K_p$ of $\mathbb Q_p$, and thus arises from a class in $\mathrm{SO}_3( K_p)/ \mathrm{SO}_3(\mathcal O_{K_p})$ where $\mathcal O_{K_p}$ is the ring of integers.
The set $\mathrm{SO}_3( K_p)/ \mathrm{SO}_3(\mathcal O_{K_p})$ is countable by a standard argument - it is a countable union of the compact sets of matrices with bounded negative valuations, and because $\mathrm{SO}_3(\mathcal O_{K_p})$ is open, the topology is discrete.
Because there are countably many extensions $K_p$, the set is countable in total.