Is $\sum_{n\leq x}{z^{\Omega(n)}} = O(x^{\frac12 + \varepsilon})$ equivalent to the Riemann hypothesis for all roots of unity $z\neq1$?

Question

$\Omega(n)$ is the number of prime divisors of $n$, counted with multiplicity.

For $z=-1$, $z^{\Omega(n)} = \lambda(n)$ is the Liouville function, and it's known that $\sum_{n\leq x}\lambda(n) = O(n^{\frac12 + \varepsilon})$ is equivalent to the Riemann hypothesis.

Is this the case for all roots of unity other than 1? Is this a known result?

If we look at the Dirichlet series $F_z(s) = \sum_{n=1}^{\infty} \frac{z^{\Omega(n)}}{n^s} = \prod_p {\frac{1}{1 - z p^{-s}}}$, we have for the product of roots of unity of order $n$: $\prod_{k=1}^n F_{e^{\frac kn2\pi i}}(s) = \prod_p {\frac{1}{1 - p^{-n s}}} = \zeta(n s)$, and using Mobius inversion we can look only at the primitive roots of unity of order $n$, $\prod_{k\leq n\\ (k,n)=1} F_{e^{\frac kn2\pi i}}(s) = \prod_{d | n} \zeta(d s)^{\mu(\frac nd)}$, but I'm not sure how to continue from here and treat each of the roots separately.

Heuristically, $\Omega(n)$ is normal with increasing mean and variance, so for any order of a root of unity $k$ I'd expect $\Omega(n) \mod k$ to be uniform as $n\to \infty$, so $\sum z^{\Omega(n)}$ should be approximately a random walk in 2D and thus $O(n^{\frac12 + \varepsilon})$.

Answer 1

Summarizing the discussion in the comments ($+\epsilon$ more):

For roots of unity other than $-1$, one does not get square-root cancellation. In fact, an application of the Selberg-Delange method (see, for example, Chapter 5 of Tenenbaum's book) gives that if $z \in S^1 \setminus \{-1\}$, then $$\sum_{n\leqslant X} z^{\Omega(n)} \sim \frac{H_z(1)}{\Gamma(z)}\cdot\frac{X}{(\log X)^{1-\Re z}} , (\star)$$ where $$H_z(s) = \zeta(s)^{-z} F_z(s),$$ is an Euler product that is convergent (and hence holomorphic) for $\Re(s) > 1/2$.

Note that this does not actually contradict the equidistribution of $\Omega(n)$ modulo $k$ for any $k$ which you alluded to in your question, since Weyl's criterion tells you that such equidistribution is equivalent to the assertion that $$\sum_{n\leqslant X} e\big(\frac{a}{k}\Omega(n)\big) = o(X),$$ for $k \nmid a$ and $e(t) = e^{2\pi it}$, which follows from $(\star)$ with $z = e(a/k)$ since the real part of a nontrivial root of unity is strictly less than $1$.

Finally, see this question and its responses, which is basically about the same question. In particular, see this paper linked in so-called friend Don's answer, which seems to be the first in the literature to address your question.