Given any non-real, complex number with positive real part, must some power of it has negative real part?

Yes, it's true.

Without loss of generality, we focus on $z$ with $|z| = 1$, and write $z = \cos(\theta) + i \sin(\theta)$, with $\theta \in (0, \frac\pi2) \cup ( \frac{3\pi}{2}, 2\pi)$ (since you're excluding real $z$, and only allowing $z$ with positive real part). Recall that by De Moivre, we have $z^k = \cos(k \theta) + i \sin(k \theta)$.

Note that the complex conjugate $\bar{z}$ has the same real part as $z$, and the same goes for $\bar{z}^n = \overline{z^n}$. For this reason, $z$ has some power with negative real part if and only if $\bar{z}$ does. Thus, we may further suppose that $\theta \in (0, \frac\pi2)$.

The basic idea is that $\theta$ must be between two angles of the form $\frac\pi n$ and $\frac{\pi}{n+1}$ for large enough $n$, and these are (relatively) easily shown to have a common multiple in the 2nd quadrant, hence some multiple of $\theta$ is there too.

Because $\theta \in (0, \frac\pi2)$, there exists some $n$ so that $\frac{\pi}{n+1} \le \theta < \frac{\pi}{n}$ (this is equivalent to an $n$ such that $n < \frac{\pi}{\theta} \le n + 1$), where $n > 2$.

Consequently, $\frac{n}{n+1}\pi \le n \theta < \pi$, where $$\frac{n}{n+1}\pi = (1 - \frac{1}{n+1})\pi > (1 - \frac{1}{n})\pi > \frac\pi2$$ since $n > 2$.

Thus, $\frac{\pi}{2} < n\theta < \pi$, hence $\cos(n\theta)$, the real part of $z^n$, is negative.

If you think in terms of Eulers formula and powers as rotation, then the answer is obviously yes. For some rotation, the complex number will end up in the 2nd or 3rd Quadrant.

You are sure to have $0<x/|z|<1$. Calculate $\theta=\cos^{-1}(x/|z|)$ which will be greater than zero, then define $n$ as the largest whole number for which $n\theta \le \pi/2$. What is $z^{n+1}?$