Here, $p$ is the canonical projection of $SL(n,\mathbb{C}) \to SL(n,\mathbb{C})/C_n$. It suffices to show that $\tilde\varphi$ is similar to a unitary representation. Explicitly, one has

$$H=\{(g,A) \in G \times SL(n,\mathbb{C}):p(A)=\varphi(g)\},$$

with $\tilde\varphi:(g,A) \mapsto A$ and $\tilde{p}:(g,A) \to g$. Since $G$ is compact and $\tilde{p}$ has finite kernel $C_n$, one sees that $H$ is a compact Lie group. Therefore the matrix representation $\tilde\varphi:H \to SL(n,\mathbb{C})$ is similar to a homomorphism $H \to SU(n)$, from which the lemma follows. $\square$

Therefore we are reduced to considering homomorphisms

$$\varphi:SO(3) \to SU(n)/C_n$$

for sake of this post. But we are not done yet. Having to deal with a quotient group is not satisfactory anyway.

Since $SU(n)$ is simply connected (see this video), the projections $SU(n) \to SU(n)/C_n$ are universal coverings. In particular, when $n=2$, we see $SU(2) \to SU(2)/C_2 = SO(3)$ is our well-known universal covering. If we lift $\varphi$ to universal coverings, we see ourselves dealing with $SU(2) \to SU(n)$. To be precise, we have the following commutative diagram (universal cover is a functor):

Dealing with $\tilde\varphi$ is much simpler. Physicists are more interested in unitary representations of the quaternion group $SU(2) = \operatorname{Spin}(3)$ rather than $SO(3)$, even though it looks more natural.