The answer to both of your questions is negative, even if you replace $q^{-n}$ by any other sequence $(a_n)$ of natural numbers.

The reason is, broadly speaking, that you may have of course have maps having multiple periodic points, and under small perturbations these will yield maps having a cycle of potentially large period nearby.

A little bit more precisely, let us consider the family of quadratic polynomials, $f_c(z) = z^2+c$. The following is well-known:

**Proposition.** Suppose that $f_c$ has a periodic point $z_0$ of period $n$, with corresponding multiplier $\mu = (f_c^n)'(z_0) = e^{2\pi i p/q}$, for some $p,q\in\mathbb{N}$.

Then, for any neighbourhood $U$ of $z_0$, there is a parameter $c'$ (arbitrarily close to $c$) such that $f_{c'}$ has a periodic point $z'$ of period $nq$ such that $f_{c'}^{kn}(z')\in U$ for $k\geq 0$, and such that $z'$ itself has multiplier $\mu' = (f_{c'}^{nq})'(z') = e^{2\pi i /q'}$ for some (large) integer $q'$.

(In addition, the map $f_{c'}$ also has a (repelling) periodic point of period $n$ in $U$.)

(What happens is that the parameter $c$ is on the boundary of two components of the interior of the Mandelbrot set, one consisting of points having an attracting orbit of period $n$ and one having an attracting orbit of period $nq$. The perturbation is obtained by passing a little bit along the boundary of the latter component.)

Now start e.g. with the map $f_{1/4}$ and $z=1/2$, and apply this proposition inductively. Take a limit of the resulting parameters (taking care to make each perturbation small enough as to not destroy the features already constructed). Clearly in this way we can find a quadratic polynomial having periodic points $(z_j)_{j\geq 0}$ of periods $(n_j)$, where
$$ n_j = \prod_{k=1}^j q_j$$.
(Here $q_j$ is some rapidly increasing sequence of positive integers obtained in the construction.)

Given any sequence $(\varepsilon_n)$, we can inductively carry out the construction so that $q_j$ points of the orbit of $z_j$ are within distance $\varepsilon_{n_j}$ of each other. Letting $\varepsilon_n$ tend to zero faster than $a_n$, the claim follows.