The answer to both questions is negative. I will now try to be careful about constants.

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

By Erdős–Kac, $\frac{1}{N} \sum_{n \leq N} a_n = o(1) $.

On the other hand, by Erdős–Kac applied to $n/q$,  $$ \frac{1}{N/q} \sum_{n \leq N \colon q\mid n} a_n = o(1) + 2 \int_0^{ \frac{\omega(q)}{\sqrt{\log \log N}}} \frac{1}{\sqrt{2\pi}}e^{ -x^2/2} dx $$ 


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 $\epsilon $  as long as $$\frac{\omega(q)}{\sqrt{\log \log N}} > \sqrt{2} \operatorname{erf}^{-1} (\epsilon) =  \sqrt{\frac{\pi}{2}} \epsilon + O(\epsilon^2)$$

So it is possible for the discrepancy to be larger than $\epsilon$ for all $q$ with more than $(1+o(1) \sqrt{\frac{\pi}{2}} \epsilon \sqrt{\log \log N}$ prime factors.