THE ANSWER IS NO.
Page numbers are from Boundary Behavior of Conformal Maps by Christian Pommerenke
The Pommerenke book concentrates on the unit disk, so here is some terminology for the theorem:
(page 1): he writes $\mathbb D$ for the open unit disk, then $\bar{\mathbb D}$ for the closed unit disk, finally $\mathbb T$ for the unit circle.
(page 30): A bounded simply connected domain $G$ in the complex plane comes equipped with a set called its prime ends, the set denoted $P(G).$
(page 32): For a prime end $p \in P(G),$ there is a set $I(p) \subseteq \mathbb C$ called the impression of $p.$ The impression of a prime end is a non-empty compact connected set.
(page 30): Theorem 2.15. Let f map $\mathbb D$ conformally onto the bounded simply connected domain $G,$ by the Riemann Mapping Theorem. There is an induced bijection $\hat{f}$ with $ \hat{f} : \mathbb T \rightarrow P(G). $
(page 35): Corollary 2.17(ii) with $\zeta \in \mathbb T,$ the mapping $f$ extends continuously to $\zeta,$ with limit $a,$ if and only if $ I( \hat{f}(\zeta)) = \{a\}, $ the singleton set consisting of the point $a.$
(page 35): Exercise 2.5.2. Let $ G = \mathbb D \backslash \{ (1 - e^{-t})e^{it} : 0 \leq t < \infty \}. $ Then $G$ has a prime end $p$ with $I(p) = \mathbb T.$ That is, $I(p)$ is not a single point.
Thus, there is some point $\zeta \in \mathbb T,$ with $\zeta = \hat{f}^{-1}(p), $ at which $f$ does not extend continuously. Put another way, $f(\zeta)$ cannot be defined in a way that makes $f$ continuous at $\zeta.$
To return to the upper half plane, let us call it $\mathbb U.$ Take a Möbius transformation $m$ that maps the real line to $\mathbb T$ minus one point of course, does NOT map $\infty$ to the point $\zeta$ where $f$ is discontinuous, and maps $\mathbb U$ conformally onto $\mathbb D.$ Then the function composition $ f \circ m : \mathbb U \rightarrow G $ and is thus bounded, but cannot be extended continuously at $m^{-1}(\zeta)$ on the real line.