I think the answer to both questions is negative.

Let $\omega(n)$ be the number of prime factors of $n$, let $C$ be a large constant, and let $$a_n = \begin{cases} 1 &  \omega(n) <  \log \log N  + C \sqrt{\log \log  N} \\ -1 & \omega(n) > \log \log N  + C \sqrt{\log \log  N} \end{cases}. $$

By Erdős–Kac, $$\frac{1}{N} \sum_{n \leq N} a_n> 1- \delta_C$$ with $\delta_C$ going to $0$ as $C$ grows. 

Optimizing this argument and the argument you mention in your answer could possibly produce a sharp

On the other hand, by Erdős–Kac applied to $n/q$,  $$ \frac{1}{N/q} \sum_{n \leq N \colon q\mid n} a_n <-1 + \delta_C$$ with $\delta_C$ going to $0$ as $C$ grows as long as $\omega(q) > 2 C \sqrt{ \log \log N}$.

So the difference $\frac{1}{N/q} \sum_{n \leq N \colon q\mid n} a_n  - \frac{1}{N}  \sum_{n\leq N }a_n $ is at least $2-2 \delta_C$  as long as $\omega(q) > 2 C \sqrt{ \log \log N}$.

By Erdős–Kac one last time, this property holds for almost all $q$ greater than $e^{ e ^{ 2 C \sqrt{\log \log N}}}$ and $q$ below that cutoff contribute only $e ^{ 2 C \sqrt{\log \log N}} \approx e^{2 C \sqrt{ \log \log Q}}$, which is much smaller than $\log Q$, to the sum.

It's interesting that this quantity looks similar to the upper bound you suggest in your answer can be obtained by pushing the argument using the theorem of Bombieri. Maybe optimizing both arguments gives lower and upper bounds that are somewhat close together, although I don't know what happens with $q>e^{ e ^{ 2 C \sqrt{\log \log N}}}$, which are easy to ignore if the goal is to only get $o(\log Q)$.