Let $D^+ = \{z \in \mathbb{C}: |z|<1 \ \mbox{and} \ \mbox{Im}(z)>0 \}.$
Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a continuous function on $\overline{D^+}$, and $f$ is holomorphic on $D^+$ satisfying $$(1) \ \ f(-x) \in \mathbb{R} \ \mbox{for} \ x>0, \ \ (2) \ \ if(x) \in \mathbb{R} \ \mbox{for} \ x>0.$$ Then $$\frac{|f(z)|}{|z|^{1/2}}$$ is bounded on $D^+.$
$\textbf{Attempt}$ So $f$ is continuous on $\overline{D^+}$, holomorphic on $D^+$ and
1) $f$ is real-valued on the left side of $x-$axis
2) $f$ is purely imaginary on the right side of $x-$axis.
Set $f(z) = u(z) + iv(z)$. Then by continuity at $0$ $$f(0) = \lim_{z \rightarrow 0,x<0 } u(z) = \lim_{z \rightarrow 0,x>0 }iv(z)$$ which yields $f(0) = 0.$
The assumption regarding $f$ seems very closed to the assumption in Schwarz reflection principle theorem except $f$ might not be real-valued on the piece of $x$-axis in $\overline{D^+}.$
Not sure which way I should proceed next ...
$f^2$ satisfies the schwarz reflection so it has an analytic continuation across the reals, so in particular at $0$; then if we write $f^2(z)=zg(z)$ with $g$ analytic around zero (since as you noted, and is easy to see by continuity also, $f(0)=0$, hence $f^2(0)=0$) we immediately get $|f^2(z)| \le c|z|$ near zero, hence the required result, since away from zero it follows just by continuity