Fix a prime $p$ and let $\rho:G_{\mathbb{Q}_p}\rightarrow GL_2(\overline {\mathbb{F}}_p)$ be an arbitrary continuous representation. I found the following statement in a paper on non-ordinary modular forms: If $\rho$ is irreducible then $\rho|_{I_p}$ is tamely ramified, where $I_p$ is the inertia subgroup of $G_{\mathbb{Q}_p}$.
I honestly am not sure exactly what the definition of tamely ramified is in this setting, but since the image of $\rho$ is finite (see here), I assume it means that the order of the image is prime to $p$.
Here are some thoughts. We have the following exact sequence $$ 1\rightarrow I_{p,w}\rightarrow I_p\rightarrow I_{p,t}\rightarrow 1 $$ where the $w$ and $t$ subscripts denote the wild ramification group and the tame quotient, respectively. Hence, if (for some reason) $I_{p,w}$ were equal to the kernel of $\rho|_{I_p}$, then the image would be isomorphic to $I_{p,t}= I_p/I_{p,w}$, which has order prime to $p$. I'm not sure why this would be true though, and am not sure why irreducibility would be important here.