Edited: The previously found sufficient condition is indeed necessary, but even better, it is satisfied by all polynomials of degree at least two. Thus the conjecture is true:

Theorem: Let $p$ be a complex polynomial of degree $d \geq 2$ and let $H_p={\rm conv}J_p$, the convex hull of the Julia set $J_p$ of $p$. Then $p^{-1}(H_p) \subset H_p$.

To prove this, I use several lemmata. First, there is the following version of Gauss-Lucas theorem, due to W. P. Thurston, which can be found in the preprint

A. Ch\'eritat, Y. Gao, Y. Ou, L. Tan: \emph{A refinement of the Gauss-Lucas theorem (after W.
P. Thurston)}, 2015, preprint, hal-01157602

Lemma 1: Let $p$ be any polynomial of degree at least two. Denote by $\mathcal{C}$ the convex hull of the critical points of $p$. Then $p: E \to \mathbb{C}$ is surjective for any closed half-plane $E$ intersecting $\mathcal{C}$.

From this we have the following:

Lemma 2: Let $p$ be any polynomial of degree at least two. Then all zeros of $p'$ belong to $H_p={\rm conv}J_p$, the convex hull of the Julia set $J_p$ of $p$.

Proof: Suppose there is an $x_0 \not \in H_p$ such that $p'(x_0)=0$. By the hyperplane separation theorem , there exists a closed half-plane $E$ such that $x_0 \in E$ and $E \cap H_p = \emptyset$. In particular, $E \cap J_p = \emptyset$. By Lemma 1, $p: E \to \mathbb{C}$ is surjective. Take a $z_0 \in J_p$. Then on one hand $p^{-1}(z_0) \subset J_p$, while on the other hand $p^{-1}(z_0) \cap E \neq \emptyset$, a contradiction.

The next lemma is a modification of Exercise 2.1.15 in
Lars Hörmander: Notions of Convexity
Publisher Springer Science \& Business Media, 2007 (Modern Birkhäuser Classics)
ISBN 0817645853, 9780817645854

Lemma 3:
Let $p(z)=\sum_{j=0}^d a_jz^j$ be a polynomial in $z \in \mathbb{C}$ of degree $d$. Let $B$ be a closed convex subset of $\mathbb{C}$ containing all zeros of $p'$. Then the set $C_B$ of all $w \in \mathbb{C}$ such that all the zeros of $p(\cdot)-w$ are contained in $B$ is a convex set.

Proof of Lemma: Note that by continuity of roots $C_B$ is closed when $B$ is. Let $w_1,w_2 \in C_B$ and $n_1,n_2 \in \mathbb{N}$ and consider the polynomial (in one complex variable $z$) $P(z):=(p(z)-w_1)^{n_1}(p(z)-w_2)^{n_2}$. Then all zeros of $P$ lie in $B$ (by definition of $C_B$), so the convex hull of zeros of $P$ is contained in $B$. By Gauss-Lucas theorem (standard version), all zeros of $P'$ are contained in $B$. The zeros of $P'$ are respectively all the zeros of $p(z)-w_1$, all the zeros of $p(z)-w_2$ (if $n_1, n_2 >1$), all the zeros of $p'$ and all the zeros of $p(\cdot)-\left (\frac{n_1}{n_1+n_2}w_1 + \frac{n_2}{n_1+n_2}w_2\right)$. By definition of $C_B$, $\frac{n_1}{n_1+n_2}w_1 + \frac{n_2}{n_1+n_2}w_2 \in C_B$. Varying $n_1,n_2$ and using the property that $C_B$ is closed, we get that $tw_1+(1-t)w_2 \in B$ for all $0 \leq t \leq 1$.

Proof of Theorem: Applying the Lemma 3 to $B=H_p={\rm conv}J_p$, we get that the set $C_p=\{w \in \mathbb{C}: \{z : p(z)-w=0\} \subset H_p\}$ is convex. Furthermore, for $w \in J_p$ all solutions of $p(z)-w$ are in $J_p \subset H_p$, so $J_p \subset C_p$. Hence $H_p \subset C_p$, which implies that $p^{-1}(H_p) \subset H_p$.

For the quadratic family $f_c(z)=z^2+c, \ c \in \mathbb{C}$ it is straightforward (without appealing to Lemma 2) to check that the critical point $0$ is the center of symmetry of the Julia set $J_c$, so it is a convex combination of two points in $J_c$.