= Solution
On the small clockwise <contour> $z=\epsilon e^{i\theta}$, $\theta$ decreases from $\pi$ to $-\pi$. Uniformly on that arc, $e^{z-\sqrt z}=1+O(\sqrt\epsilon)$, and hence
$$
\int_{C_4}F(z)\,dz=i\int_\pi^{-\pi}\bigl(1+O(\sqrt\epsilon)\bigr)\,d\theta\longrightarrow-2\pi i.
$$
The large arcs also vanish. To make this precise for the displayed geometry, take them on the left semicircle $z=\gamma+Re^{i\theta}$, $\pi/2\le\theta\le3\pi/2$, split at the cut. On the principal <holomorphic branch>, $\operatorname{Re}\sqrt z\ge0$, $|z|\ge R-\gamma$, and $\operatorname{Re}z=\gamma+R\cos\theta$. The <estimation lemma> gives
$$
\left|\int_{C_2\cup C_6}F\,dz\right|\le\frac{Re^\gamma}{R-\gamma}\int_{\pi/2}^{3\pi/2}e^{R\cos\theta}\,d\theta=O(R^{-1}).
$$
For the last bound, symmetry and $\sin u\ge2u/\pi$ on $0\le u\le\pi/2$ imply that the angular integral is at most $\pi/R$. The exact choice of large arcs does not affect the limiting result. The vertical <improper integral> converges absolutely because $e^{-\operatorname{Re}\sqrt{\gamma+iy}}/|\gamma+iy|$ decays as $O(e^{-c\sqrt{|y|}}/|y|)$.
Taking limits in the closed <contour integral> now yields $2\pi iJ+2iK-2\pi i=0$, where $K=\int_0^\infty e^{-x}\sin(\sqrt x)\,dx/x$. Thus
$$
\boxed{J=1-\frac1\pi\int_0^\infty e^{-x}\frac{\sin\sqrt x}{x}\,dx.}
$$
As an optional explicit evaluation, put $x=t^2$ and define $K(a)=2\int_0^\infty e^{-t^2}\sin(at)\,dt/t$. <Differentiation under the integral sign> and the <Gaussian integral> give $K'(a)=\sqrt\pi e^{-a^2/4}$ and $K(0)=0$. Therefore $K=K(1)=\pi\operatorname{erf}(1/2)$ and $J=\operatorname{erfc}(1/2)\simeq0.479500$ in terms of the <error function> and <complementary error function>.
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