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.
So like you said you want to find $p,c_4,c_6$ such that (assuming $p\ne 2,3$)
- $c_4\ne0$ and $c_6 \ne 0$
- $c_4^3 -c_6^2 \ne 0$
- $p|(c_4^3 -c_6^2)$
- $3v_p(c_4)=v_p(c_4^3)\ge v_p(c_4^3 -c_6^2)$ (integrality of $j$)
One way to get 3. to hold is for both $c_4,c_6$ to be divisible by $p$.
If $3v_p(c_4) =v_p(c_4^3) > v_p(c_6^2)=2v_p(c_6)$ then $v_p(c_4^3 -c_6^2) = v_p(c_6^2)$ so we get 4. for free.
So how to get $3v_p(c_4) > 2v_p(c_6)$ but $v_p(c_4)\ge 1$ and $v_p(c_6)\ge 1$, we can just take both to have valuation 1!
So why not try $c_4 = c_6 = p$, this satisfies all properties!
So for example $E : y^2 = x^3 -27\cdot5 x -54\cdot 5$ which is https://www.lmfdb.org/EllipticCurve/Q/10800dg1/
Best Answer
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...