Choose $g(z) = z^5 + 1$. The zeros of $g$ are $e^{k\pi i/5},\; k \in \{1,3,5,7,9\}$, two of which lie in the right half plane.
Then a slightly generalised version of Rouché's theorem tells you that $h$ also has two zeros in the right half plane.
The standard formulation of Rouché's theorem demands that $\lvert h-g\rvert < \lvert h\rvert$ (or $\lvert h-g\rvert < \lvert g\rvert$) on the boundary $\partial V$ of the region, but looking at the proof, whose main part is the observation that the number of zeros of $f_\lambda = g + \lambda(h-g)$ in $V$,
$$N_\lambda = \frac{1}{2\pi i}\int_{\partial V} \frac{g'(z) + \lambda\bigl(h'(z) - g'(z)\bigr)}{g(z) + \lambda\bigl(h(z) - g(z)\bigr)}\, dz$$
is independent of $\lambda \in [0,\,1]$, since no $f_\lambda$ has a zero on $\partial V$.
The role of the condition $\lvert h-g\rvert < \lvert g\rvert (\text{ or } \lvert h\rvert)$ on $\partial V$ is solely to guarantee that $g + \lambda(h-g)$ also has no zeros on $\partial V$. If we can determine that in a different way, the conclusion still holds.
For a large enough semicircle in the right half plane, we have $\lvert 5z^3 + 2z^2 + 4z\rvert < \lvert z^5 + 1\rvert$ since the RHS is a polynomial of higher degree than the LHS.
On the imaginary axis, although $\lvert g-h\rvert > \lvert g\rvert$ on some parts, expanding shows
$$g(it) + \lambda\bigl(h(it) - g(it)\bigr) = 1 + it^5 + \lambda(-5it^3 - 2t^2 + 4it) = (1-2\lambda t^2) + it(t^4 - 5\lambda t^2 + 4\lambda),$$
and the real part vanishes only for $\lambda = \dfrac{1}{2t^2}$ (with $t^2 \geqslant \frac12$, since $\lambda \leqslant 1$), when the imaginary part is
$$t\left(t^4 - \frac{5}{2} + \frac{2}{t^2}\right) = \frac1t\left(t^6 - \frac{5}{2}t^2 + 2\right),$$
and the polynomial $x^3 - \frac52x + 2$ has no positive real roots, so we conclude $g(z) + \lambda(h(z)-g(z)) \neq 0$ on the imaginary axis.
Best Answer
Some analysis of the behavior on the imaginary axis allows you to tell the net change in argument over the diameter of the semicircle, from $Ri$ to $-Ri$ for some big positive $R$. Noticing that the real part of $f(it)$ is $t^4-3t^2+3=\left(t^2-\frac{3}{2}\right)^2+\frac{3}{4}>0$, you can see first of all that there can be no winding about zero on the imaginary axis. Furthermore, since the real part of $f$ on the imaginary axis has degree $4$ and the imaginary part of $f$ on the imaginary axis has degree $3$, the real part of $f(\pm Ri)$ will be much larger than the imaginary part of $f(\pm Ri)$ for large $R$, which means that the argument of $f$ at the endpoints of the diameter will be near zero, and you can conclude that there is near $0$ net change in argument along the diameter of the semicircle.
This leaves the analysis of the change in argument of $f(Re^{i\theta})$ as $\theta$ ranges from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. For this, it is helpful to note that $\frac{1}{R^4}f(Re^{i\theta})=e^{4i\theta}\left(1+\frac{8}{R}e^{-i\theta}+\frac{3}{R^2}e^{-2i\theta}+\frac{8}{R^3}e^{-3i\theta}+\frac{3}{R^4}e^{-4i\theta}\right)$ is near $e^{4i\theta}$ when $R$ is large.