Consider the set $S = \{x + iy \mid x \in \mathbb{R}-\{0\},\, y = x\sin(1/x)\} \cup \{0\} \subseteq \mathbb{C}$, which looks like the graph of $x\sin(1/x)$ lying in the complex plane. Let $B = B_{\varepsilon}(0)$ be a ball of radius $\varepsilon > 0$ around $0$.
Is there a conformal (or even holomorphic) function $f$ on $B$ so that $f(S \cap B)$ is a real interval $(a, b) \subseteq \mathbb{R}$ (with $a < b$)?
It seems like the answer should be no. In particular, given that for any $\varepsilon > 0$, there will be $x_{1}, x_{2}, x_{3} \in (0, \varepsilon)$ where $x_{1} \in \mathbb{R}$, $x_{2} \in \mathbb{H}$, and $x_{3} \in -\mathbb{H}$, it doesn't seem like it would be possible to conformally "smooth" this out into a straight line. However, I don't have any idea how to rigorously prove this.
As you expected it, the answer is no. Suppose $f$ is holomorphic on $B(0,r)$ and $f$ is real on the range of $\gamma,$ where $\gamma(t) = t+it\sin(1/t), t \in (0,r).$ We will show $f$ is constant in $B(0,r).$
Note that $f(1/(2\pi n))$ is real for large $n.$ Writing $f=u+iv,$ this implies $v(1/(2\pi n))=0$ for large $n.$ Now $v$ is real analytic on $(-r,r),$ so by the indentity principle, $v=0$ on $(-r,r).$ Hence $f$ is real on $(-r,r).$
For large $n,$ the closure of the region
$$U=\{x+iy: 1/(2\pi(n+1))<x<1/(2\pi n), 0< y < x\sin(1/x)\}$$
is contained in $B(0,r).$ Fix such an $n.$ Then $f$ is real on $\partial U,$ i.e., $v=0$ on $\partial U.$ By the maximum principle for harmonic functions, $v=0$ in $U.$ Thus $f$ is real valued in $U,$ hence $f$ is constant in $U$ by the open mapping theorem. By the identity principle, $f$ is constant in $B(0,r)$ as claimed.