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 $D_r(\vec v)$ the open radius-$r$ disc centered at $\vec v$, where $r=1$ if it's omitted.

> **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 $\gamma(a), \gamma(c) \notin \bar D(\gamma(b))$, 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 $\vec v \neq \vec w$.
We say a set $K \subset \mathbb{R}$ is *between* two parallel lines $\ell$ and $\ell'$, if $K$ is contained in the convex hull of $\ell \cup \ell'$.

> **Lemma 3.**
> There exist two parallel lines such that $\gamma(\mathbb{R})$ is between them.
>
> *Proof.*
> For $x > 0$, consider $C_x = \gamma([-x,x])$.
> Since $C_x$ is compact, it is between two lines $\ell_x^-, \ell_x^+$ that are parallel to $\ell_x = \ell(\gamma(-x), \gamma(x))$ and have minimal distance among such pairs of lines.
> Then both lines intersect $C_x$.
>
> We claim that the distance between $\ell_x^-$ and $\ell_x^+$ is at most $2$.
> If not, one of them, say $\ell_x^+$, has distance $d > 1$ from $\ell_x$.
> Take any $\gamma(b) \in \ell_x^+ \cap C_x$, which has distance at least $d$ from $\gamma(-x)$ and $\gamma(x)$, since the latter points are in $\ell_x$.
> Then the numbers $-x < b < x$ and the line $\ell_x^+$ contradict Lemma 2.
>
> We claim that as $x \to \infty$, the direction of the line $\ell_x$ converges to some direction $\delta$.
> For all $0 < x < y$ we have $\gamma(-x), \gamma(x) \in C_y$, and hence the segment $S_x = \gamma(-x) \gamma(x)$ of $\ell_x$ is between $\ell_y^-$ and $\ell_y^+$.
> Then every point of $S_x$ is within distance $2$ of $\ell_y$, which constrains the direction of $\ell_y$ to an interval whose length approaches $0$ as the length $\| \gamma(-x) - \gamma(x) \|$ of $S_x$ grows.
> The intersection of these intervals is a singleton $\{\delta\}$.
>
> Let $\ell^-, \ell^+$ be the two lines with direction $\delta$ that are at distance $3$ from $\gamma(0)$.
> We claim that $\gamma(\mathbb{R})$ is between these lines.
> Fix $x > 0$, and consider $r > 0$.
> For all large enough $y$, the set $D_r(\gamma(0)) \cap (\ell_y^- \cup \ell_y^+)$ is between $\ell^-$ and $\ell^+$ since the directions of $\ell_y^-, \ell_y^+$ converge to $\delta$ and they are at distance at most $2$ from $\gamma(0)$.
> Since $C_x$ is between $\ell_y^-$ and $\ell_y^+$, if we chose $r$ large enough then this shows that $C_x$ is also between $\ell^-$ and $\ell^+$.
> Since this holds for all $x$, we are done.
> QED.

We now fix two parallel lines $\ell^-, \ell^+$ that have $\gamma(\mathbb{R})$ between them and whose distance is minimal.

> **Lemma 4.**
>  $\ell^- = \ell^+$.

> *Proof.*
> Suppose not.
> Both lines contain points that are arbitrarily close to $\gamma(\mathbb{R})$ since their distance is minimal.
> If either line intersects $\gamma(\mathbb{R})$, then the intersection point $\gamma(b)$ and the points $\gamma(-x), \gamma(x)$ for any large enough $x > 0$ contradict Lemma 2.
> 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 $C_a = \gamma([-a,a])$ for some $a$ with $\gamma(\mathbb{R}) \cap ([-2,2] \times \mathbb{R}) \subset C_a$ and let $\epsilon > 0$ be its minimum distance from $\ell^+$.
> Take $b, c$ with $\gamma(b)_1 > 2$, $\gamma(c)_1 \gg 4 \cdot \gamma(b)_1$ 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 $C_a$.
> 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 $\ell'$ that lies above $\gamma([-a,c])$.
> Let $\gamma(b)$ be an intersection point of $\ell'$ and $\gamma([-a,c])$.
> It is easy to see that $\gamma(0), \gamma(c) \notin \bar{D} (\gamma(b))$, so that $\ell'$ violates Lemma 2 with $0$, $b$ and $c$.
> QED.

Now $\gamma$ is the line $\ell^- = \ell^+$.