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Edited question to restrict to large primes
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Given a point $P$ on a genus-$1$ curve over $\mathbb{Q}_p$, is there an $R$ such that $2 \mid [P - R]$ and $x(\overline{P}) \neq x(\overline{R})$?

Let $p \in \mathbb{Z}$ be prime, and let $f \in \mathbb{Z}_p[x]$ be a quartic polynomial with nonzero discriminant. Let $C/\mathbb{Q}_p$ be the genus-$1$ curve with affine equation $y^2 = f(x)$. Let $\overline{C}/\mathbb{F}_p$ denote the mod-$p$ reduction of $C$, and for any $P \in C(\mathbb{Q}_p)$, let $\overline{P} \in \overline{C}(\mathbb{F}_p)$ denote the mod-$p$ reduction of $P$.

Suppose that $C(\mathbb{Q}_p) \neq \varnothing$, and fix $O \in C(\mathbb{Q}_p)$. Let $E$ be the Jacobian of $C$, and let $\iota \colon C(\mathbb{Q}_p) \to E(\mathbb{Q}_p)$ be the map sending $P \mapsto [P - O]$.

Question: Suppose $p \gg 1$. Let $P \in C(\mathbb{Q}_p)$ be integral, meaning that $x(\overline{P}) \neq \infty$. Does there exist $Q \in C(\mathbb{Q}_p)$ such that $\iota(P) + 2 \cdot \iota(Q) = \iota(R)$, where $R \in C(\mathbb{Q}_p)$ is integral and $x(\overline{P}) \neq x(\overline{R})$?

Partial Answer: Suppose that $C$ has good reduction modulo $p$ (i.e., suppose that $p$ does not divide the discriminant of $C$), and let $\overline{E}/\mathbb{F}_p$ denote the mod-$p$ reduction of $E$. Then $\#\overline{C}(\mathbb{F}_p) \gg p$ by the Hasse bound, so the set of points of the form $\iota(\overline{P}) + 2 \cdot \iota(\overline{Q}) \in \overline{E}(\mathbb{F}_p)$ has size $\gg p$. But the set of points $\overline{R} \in \overline{C}(\mathbb{F}_p)$ such that $x(\overline{R}) \in \{\infty, x(\overline{P})\}$ has size $\ll 1$, so the answer to the question in this case is yes if $p$ is sufficiently large. I'm not sure how to make the above argument work in the case where $E$ does not have good reduction modulo $p$, because I can't talk about the group $\overline{E}(\mathbb{F}_p)$.

Edits: I've edited the question to include the assumption $p \gg 1$, so as to avoid the issue mentioned by Chris Wuthrich, which is that $\overline{C}(\mathbb{F}_p)$ can be $2$-torsion.

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