I believe the implication is true for an arbitrary simple curve $\gamma$ whose both ends escape to infinity. First, let's rescale so that the discs have radius $1$. Denote by $D(\vec v)$ the open unit disc centered at $\vec v$.
Lemma 1. For each $x$, $\gamma^{-1}(D(\gamma(x)))$ is an interval.
Proof. If not, then it has at least two connected components, whose $\gamma$-images split $D(\gamma(x))$ into at least three components. QED.
The following lemma does most of the work.
Lemma 2. Let $a < b < c$ be such that $a, c \notin \bar D(\gamma(b))$, and let $\ell$ be a line passing through $\gamma(b)$. If $\gamma([a,c])$ is not contained in $\ell$, then it intersects both components of $\mathbb{R}^2 \setminus \ell$.
Proof. Suppose not, and let $A, B$ be the components with $\gamma([a,c])$ intersecting $B$. Re-orient the plane so that $\ell$ is horizontal. Let $x = \sup \{ t \leq b : \gamma(t) \in B \}$ or $x = \inf \{ t \geq b : \gamma(t) \in B \}$, whichever is finite and minimizes $\| \gamma(x) - \gamma(b) \|$; WLOG assume the latter and that $\gamma(x)$ is on the right of $\gamma(b)$ on $\ell$. Then there exist $t < x$ with $\gamma(t) \in B$ arbitrarily close to $x$. For $r > 0$, let $\ell_r$ be the length-$r$ closed segment of $\ell$ that extends to the right from $\gamma(x)$. Then $\gamma(\mathbb{R})$ cannot contain $\ell_r$ as a subset: since $\gamma$ is a homeomorphism onto its image, $\gamma^{-1}(\ell_r)$ would have to be connected, which contradicts the properties of $x$. If $\gamma(x) \in D(\gamma(b))$, then there exists $\vec v \in (\ell_r \cap D(\gamma(b))) \setminus \gamma(\mathbb{R})$, which has a neighborhood $U$ disjoint from $\gamma([a,b])$, and then $B \cup U$ is contained in one component of $D(\gamma(b))$ and covers more than half of it.
Suppose then $\gamma(x) \notin D(\gamma(b))$ and let $y = \inf \{ y \geq b : x \in D(\gamma(y)) \}$. Note that $S = \gamma(\mathbb{R}) \cap D(\gamma(y))$ is a line segment. If $\| \gamma(c) - \gamma(y) \| > 1$, then $c+\epsilon$ for a small $\epsilon$ works as $b$ above. Otherwise we must have $\gamma(c) \in B$, since $S = \gamma((t,x)) = \ell \cap D(\gamma(y))$ for some $t$ and $\gamma$ is simple. The minimal $d > c$ with $\gamma(d) \in \ell$, if it exists, satisfies $\| \gamma(d) - \gamma(x) \| > 0$, so we have $\gamma(x) \in D(\gamma(y+\epsilon))$ and $\gamma(d) \notin D(\gamma(y+\epsilon))$ for some small $\epsilon > 0$. Then $y+\epsilon$ and $d$ work as $b$ and $c$ above. QED.
In the following $\ell(\vec v, \vec w)$ is the line that passes through $\vec v \neq \vec w$.
Lemma 3. There exist two parallel lines such that $\gamma(\mathbb{R})$ is contained in their convex hull.
Proof. For $x > 0$, consider the two lines $\ell_x^-, \ell_x^+$ that are parallel $\ell(\gamma(-x), \gamma(x))$, contain $\gamma([-x,x])$ in their convex hull, and have minimal distance. Both lines intersect $\gamma([-x,x])$. If the distance between the lines is greater then $2$, one of them contradicts Lemma 2. As $x \to \infty$, the lines converge to some parallel lines $\ell^-, \ell^+$ with distance at most $2$ whose convex hull contains $\gamma(\mathbb{R})$. QED.
The final part is a bit handwave-like, but if you draw a picture the idea should be clear, and this answer is already pretty long.
Lemma 4. $\ell^- = \ell^+$.
Proof. Suppose not. Both lines contain points that are arbitrarily close to $\gamma(\mathbb{R})$, and by Lemma 2 neither can intersect it. Rotate the plane so the lines are horizontal with $\ell^+$ on top, and suppose WLOG $\gamma(x)$ escapes to the right when $x \to +\infty$ and gets arbitrarily close to $\ell^+$ as it does so. Consider $K = \gamma([-a,a])$ for some $a$ with $\gamma(\mathbb{R}) \cap ([-2,2] \times \mathbb{R}) \subset K$ and let $\epsilon > 0$ be its minimum distance from $\ell^+$. Take $b, c$ with $\gamma(b)_1 > 2$, $\gamma(c)_1 \gg 2 \cdot \gamma(b)_1$ and both $\gamma(b), \gamma(c)$ closer to $\ell^+$ than $\epsilon/2$. Consider the minimum-slope line $\ell$ that passes through $\gamma(c)$ and intersects $K$. The vertical distance from $\gamma(b)$ to $\ell$ is at least $\epsilon/4$, and so is the vertical distance between $\ell$ and the lowest parallel line that lies above $\gamma([-a,c])$. This line violates Lemma 2 with $0$ and $c$. QED.
Now $\gamma$ is the line $\ell^- = \ell^+$.