This can't be the case for all $p$-adic fields. Indeed, start with $E_1{/\mathbb{Q}_p}$ an elliptic curve with good ordinary reduction and $E_2{/\mathbb{Q}_p}$ an elliptic curve with good supersingular reduction.
Let $K = \mathbb{Q}_p(E_1[p](\overline{K}), E_2[p](\overline{K}))$. Then upon base extension to $K$, both $E_1[p]$ and $E_2[p]$ have the same $\mathbb{F}_p[\operatorname{Gal}_K]$-module structure: namely they are both isomorphic as abelian groups to $(\mathbb{Z}/p\mathbb{Z})^2$ and both have trivial Galois action. (Moreover the ordinary/supersingular dichotomy does not change upon base extension: this depends only on the $j$-invariant of $E$ modulo $p$.)
There is something to be said in the positive direction though coming from restrictions on torsion in the formal group of $E_{/K}$ depending on the ramification index $e(K/\mathbb{Q}_p)$. Let me know if you want to hear more details about that...
An elliptic curve over $\mathbf Q$ cannot have complex multiplication (defined over $\mathbf Q$). It's possible for a rational elliptic curve to have extra endomorphisms, but these will only be defined over a finite extension.
But let's instead take an elliptic curve $E$ over a number field $K$ with complex multiplication. Then the associated Galois representation is reducible*!
Indeed, if
$$\rho_{E,\ell}:G_K\to \mathrm{GL}_2(\overline{\mathbf Q}_\ell)$$
is the associated $\ell$-adic representation, then its not too hard to check that
$$\mathrm{End}(E)\otimes_\mathbf Z\overline{\mathbf Q}_\ell\hookrightarrow\mathrm{End}(\rho_{E,\ell}).$$
In particular, if $\mathrm{End}(E)\ne \mathbf Z$, then $\mathrm{End}(\rho_{E,\ell})$ is not a field, so $\rho_{E,\ell}$ is reducible. Its subrepresentations are one dimensional Galois representations, which by class field theory, correspond to the Grossencharacters of $E$.
In fact the above map is an isomorphism (by Faltings' theorem). So if $\mathrm{End}(E) = \mathbf Z$, then $\mathrm{End}(\rho_{E,\ell})$ is a field, so $\rho_{E,\ell}$ is irreducible.
If $E$ does not have complex multiplcation over $K$, but obtains extra endomorphisms over a finite extension, then the above argument shows that $\rho_{E, \ell}$ is irreducible. However, $\rho_{E, \ell}$ will not be surjective. By Mackey theory, since $\rho_{E, \ell}$ is irreducible, but $\rho_{E, \ell}|_{G_L}$ is reducible for some $L$, we find that $\rho_{E, \ell}$ is induced from a character of a quadratic extension. In particular, its image cannot be $\mathrm{GL}_2(\mathbf Z_\ell)$.
*By reducible, I mean that it becomes reducible over the algebraic closure $\overline{\mathbf Q}_\ell$. It may still be irreducible over $\mathbf Z_\ell$.
Best Answer
Certainly. It will (typically) be irreducible for primes where the curve has supersingular reduction. Indeed, for such primes, the p-torsion as a module over the endomorphism ring $O$ is isomorphic to $O/p \simeq \mathbf{F}_{p^2}$, and the image will (for all but finitely many such primes) will be all of $\mathbf{F}^*_{p^2}$ thought of as a subgroup of $\mathrm{GL}_2(\mathbf{F}_{p})$ via the isomorphism of groups $\mathbf{F}_{p^2} \simeq (\mathbf{F}_{p})^2$ (The image will be the so-called “non-split Cartan). This is certainly always irreducible - for example, reducible subgroups have order dividing the order $p(p-1)^2$ of the Borel.
An alternative way to think about it: reducibility for almost all $p$ implies the existence of $p$ isogenies for almost all $p$. Since the isogeny class of an elliptic curve is finite, this implies that for any infinite set of primes $S$ where p-isogenies exist there will be an endomoprhism of degree $pq$ with distinct primes $p$ and $q$ in $S$. (By pigeonhole principle the two isogenies are both from E to the same E’ then take one followed by the dual isogeny of the other.) But their only exist endomorphisms of E of orders which are norms in $O$ which forces $p$ and $q$ to split in the quadratic field.