The answer is negative. Since every neighborhood of a point on the Julia set is mapped onto the whole Julia set by some iterate of the rational function, it follows that if a small piece of the Julia set is a simple curve, then the whole Julia set is either a simple curve or a Jordan curve.

Addressing the comment. Use the following Lemma. Let $g$ be a germ of an analytic function is a neighborhood of $z_0$, $w_0=g(z_0)$ and $E$ is an arbitrary set containing $w_0$. If the component of $g^{-1}(E)$ containing $x_0$ is a simple curve, then intersection of $E$ with a neighborhood of $w_0$ is either a simple curve, or a semi-closed simple curve with one end at $w_0$.

Proof. WLOG $z_0=w_0=0$, and $g(z)=z^m$. This makes the statement evident. In the first case, $m=1$, in the second case $m=2$.

It follows that the Julia set $J$ is a bordered closed 1-manifold, and thus $J$ is homeomorphic to a circle or to a segment.

The answer is negative. Since every neighborhood of a point on the Julia set is mapped onto the whole Julia set by some iterate of the rational function, it follows that if a small piece of the Julia set is a simple curve, then the whole Julia set is either a simple curve or a Jordan curve.

Addressing the comment. Use the following Lemma. Let $g$ be a germ of an analytic function is a neighborhood of $z_0$, $w_0=g(z_0)$ and $E$ is an arbitrary set containing $w_0$. If the component of $g^{-1}(E)$ containing $x_0$ is a simple curve, then intersection of $E$ with a neighborhood of $w_0$ is either a simple curve, or a semi-closed simple curve with one end at $w_0$.

Proof. WLOG $z_0=w_0=0$, and $g(z)=z^m$. This makes the statement evident. In the first case, $m=1$, in the second case $m=2$.

It follows that the Julia set $J$ is a bordered closed 1-manifold, and thus $J$ is homeomorphic to a circle or to a segment.

Edit. There is also a reference on this result: Theorem A in this paper.

The answer is negative. Since every neighborhood of a point on the Julia set is mapped onto the whole Julia set by some iterate of the rational function, it follows that if a small piece of the Julia set is a simple curve, then the whole Julia set is either a simple curve or a Jordan curve.

The answer is negative. Since every neighborhood of a point on the Julia set is mapped onto the whole Julia set by some iterate of the rational function, it follows that if a small piece of the Julia set is a simple curve, then the whole Julia set is either a simple curve or a Jordan curve.

Addressing the comment. Use the following Lemma. Let $g$ be a germ of an analytic function is a neighborhood of $z_0$, $w_0=g(z_0)$ and $E$ is an arbitrary set containing $w_0$. If the component of $g^{-1}(E)$ containing $x_0$ is a simple curve, then intersection of $E$ with a neighborhood of $w_0$ is either a simple curve, or a semi-closed simple curve with one end at $w_0$.

Proof. WLOG $z_0=w_0=0$, and $g(z)=z^m$. This makes the statement evident. In the first case, $m=1$, in the second case $m=2$.

It follows that the Julia set $J$ is a bordered closed 1-manifold, and thus $J$ is homeomorphic to a circle or to a segment.