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
tl;dr: yes for $n=1$ and $p=2$ otherwise no.
Let $L_n:=\mathbf{Q}(E[p^n])$ and $K_n:=\mathbf{Q}(\zeta_{p^n})$. We have maps: $$\mathrm{Gal}(L_n/\mathbf{Q}) \subseteq \mathrm{GL}_2(\mathbf{Z}/p^n \mathbf{Z}) \stackrel{\mathrm{det}}{\rightarrow} (\mathbf{Z}/p^n \mathbf{Z})^{\times} = \mathrm{Gal}(K_n/\mathbf{Q}),$$ and a corresponding inclusion $K_n \subset L_n$. If $L_n = L_{n+1}$, then there is a surjection:
$$\mathrm{Gal}(L_n/\mathbf{Q}) \rightarrow \mathrm{Gal}(K_{n+1}/\mathbf{Q}) = (\mathbf{Z}/p^{n+1} \mathbf{Z})^{\times}.$$
For $m \ge 1$, the group $\mathrm{Gal}(L_{m+1}/L_m)$ is a (possibly trivial) elementary group of exponent $p$. Hence it follows from the existence of the surjection above that there are also exist surjections
$$\mathrm{Gal}(L_m/\mathbf{Q}) \rightarrow (\mathbf{Z}/p^{m+1} \mathbf{Z})^{\times}$$
for every $m \le n$, and in particular for $m = 1$. We show that for $p > 2$ there are no groups $G \subset \mathrm{GL}_2(\mathbf{F}_p)$ with this property and such that the determinant map is surjective.
Since $G$ has order dividing $p$, it either contains $\mathrm{SL}_2(\mathbf{F}_p)$ (and thus is $\mathrm{GL}_2(\mathbf{F}_p)$ by the determinant assumption) or is contained in a Borel $B$.
In the first case, there are no surjective maps to a cyclic group of order $p$ when $p >2$, because $\mathrm{SL}_2(\mathbf{F}_p)$ is simple up to the center of order $2$ when $p \ge 5$, and $G = \widetilde{S_4}$ when $p = 3$ and the abelianization has order $2$.
If $G \subset B$, we may assume after conjugation that $B$ consists of upper triangular matrices and $G$ contains the element
$$\gamma = \left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right).$$
If $\beta \in B$, then the commutator $[\beta,\gamma]$ is a non-trivial power of $\gamma$ unless $\beta$ lies in the group $H$ generated by $\gamma$ and by the diagonal matrices. In particular, the only subgroups of the Borel whose commutator subgroup does not contain an element of order $p$ (which would prevent the existence of a quotient of order $p$) lie inside a group $H$ all of whose elements have determinant in $\mathbf{F}^{\times 2}_p$ which prevents the determinant map from being surjective.
This leaves the case $p = 2$. In this case, it can indeed happen that $\mathbf{Q}(E[2]) = \mathbf{Q}(E[4])$, for example when:
$$E:y^2 + x y + y = x^3 - x^2 + 4 x - 1.$$
It turns out that it is impossible that $\mathbf{Q}(E[2^n]) = \mathbf{Q}(E[2^{n+1}])$ for larger $n$, but this is a theorem of Jeremy Rouse and David Zureick-Brown that depends on showing that certain modular curves have no rational points. See Remark 1.5 of https://arxiv.org/pdf/1402.5997.pdf
The example $E$ above has a surjective $2$-torsion representation. Finally, you have the assumption "with good reduction over $\mathbf{Q}$" but this doesn't make much sense because elliptic curves by definition are smooth. If you mean "good reduction over $\mathbf{Z}$" then there are no elliptic curves at all. If you mean "good reduction at $p$" then there are also no curves; the argument above shows that $\mathbf{Q}(\zeta_{p^2}) \subset \mathbf{Q}(E[p])$, and this prevents $E[p]/\mathbf{Z}_p$ arising from a finite flat group scheme by Fontaine's bounds.