The first part of the question is easy: the power decay is typical. Let's look, say, at $\theta\in[1/3,1/2)$. Take any $\xi>0$ and consider the sequence $\nu_k=\pi^{-1}\xi\theta^k$ up to the moment it goes below $1$. Let $m$ be the number of terms in this sequence. Let $n_k$ be the nearest integer to $\nu_k$. The power decay will be ensured if we show that a noticeable portion of the differences $|\nu_k-n_k|$ are not too small for all sufficiently large $m$.
Fix $\alpha,\delta>0$ and assume that we have at most $\alpha m$ differences $|\nu_k-n_k|$ exceeding $\delta$. Let's trace the sequence $n_k$ backwards. Assume that $n_k$, $n_{k+1}$, and $n_{k+2}$ approximate the corresponding $\nu's$ with error less than $\delta$. Then we have $n_k\approx n_{k+1}^2n_{k+2}^{-1}$ with the relative error about $\delta(2n_{k+1}^{-1}+n_{k+2}^{-1})<\frac 13 n_k^{-1}$ if $\delta$ is chosen small enough. But then $n_k$ is completely determined by $n_{k+1}$ and $n_{k+2}$.
The same argument shows that even if we only know that the approximation error is at most $\frac 12$, we can have just some fixed number $A$ of choices for $n_k$.
Now, let's count the "bad" sequences of $m$ terms. We have ${m\choose \alpha m}$ ways to select the "bad approximation positions". Then we have some constant number of starting pairs $n_{m-1}, n_m$. When going backwards, we have only $3\alpha m$ places where we have any freedom. Thus, we have just $C{m\choose \alpha m}A^{3\alpha m}\le Ce^{q(\alpha)m}$ bad sequences where $q(\alpha)\to 0$ as $\alpha\to 0$. On the other hand, $n_0$ and $n_1$ determine $\theta$ with an error of order $n_1^{-1}\le 2^{-m}$. Thus, the measure of $\theta$ that are bad for some particular $m$ is exponentially small in $m$ if $\alpha$ is small enough. The rest should be clear.
The second part of the question is harder to answer. I have no doubt that people have figured out most of what would be worth figuring out here but I doubt very much they bothered to publish any of it or, if they did, the papers made it past the referees. If you need something particular, state a precise question and we'll see if we can figure it out.
Edit in response to the edit of the question.
So the rate of decay I was expecting is (or arbitrarily close to) half of the Hausdorff dimension $\frac12 dim_H(C_\theta)=\frac 12\frac{\log(1/2)}{\log(\theta)}$.
Now, that is far too optimistic. Take large $\lambda=\theta^{-1}$. Take any integer. Multiply by $\lambda$. Correct the product to the nearest integer. Multiply by $\lambda$. Correct the product to the nearest integer, and so on. After $m$ steps, you'll get $\xi\approx\lambda^m$ while the cumulative effect of all corrections in each position is at most $\theta+\theta^2+\theta^3+\dots\approx\theta$, so all the cosines are about $1-\theta^2$ and you get $\alpha$ not much better than $\frac{\theta^2}{\log(1/\theta)}$
We can find Cantor-like sets of Hausdorff dimension arbitrarily close to $1$ which satisfy your property.
Lemma: If a subset $S \subseteq \{1, 2, \dots, n\}$ of cardinality $|S| = m$ has no length-$3$ arithmetic progressions, then we can find a subset $K \subsetneq [0, 1]$ with a Hausdorff dimension of:
$$ \dfrac{\log{m}}{\log(2n-1)} $$
In particular, the Cantor ternary set is the special case of applying this construction to $S = \{1, 2\}$.
Proof: We apply a Cantor-like construction, replacing the interval $K_0 = [0, 1]$ with $m$ intervals:
$$ K_1 = \bigcup \{ [\dfrac{2s-2}{2n-1}, \dfrac{2s-1}{2n-1}] : s \in S \} $$
and iterating in the obvious way to produce a sequence $K_0 \supsetneq K_1 \supsetneq K_2 \supsetneq \dots$. Define $K$ to be the limit (intersection) of these sets.
Now given any two points $x, y \in K$, we consider the first iteration $K_n$ of the process which puts $x,y$ in different intervals. Then I claim that $\frac{x+y}{2}$ is not in $K_n$.
Due to the recursive structure of $K$, we can wlog assume $n = 1$. Hence we just need to show that if $x$ and $y$ belong to different intervals in $K_1$, their mean does not belong to $K_1$. This is where we make use of the $3AP$-less-ness of $S$.
For simplicity of notation, we'll scale $K_1$ up by $2n - 1$ and suppose that:
$$ x \in [2s - 2, 2s - 1], y \in [2t - 2, 2t - 1] $$
where $s,t \in S$ and wlog $s < t$.
Then the mean of $x$ and $y$ must consequently satisfy:
$$ \dfrac{x + y}{2} \in [s + t - 2, s + t - 1] $$
If $s + t$ is odd, this interval doesn't belong to $K_1$ by parity. If $s + t$ is even, then this interval doesn't belong to $K_1$ because $\frac{1}{2}(s + t) \notin S$ by the $3AP$-less-ness of $S$.
The result follows.
Theorem: For every $\varepsilon > 0$, we can find a set $K$ of Hausdorff dimension greater than $1 - \varepsilon$
Proof: By Lemma 1, we just need to find a $m$-subset of $[n]$ with no three-term arithmetic progression and which satisfies:
$$ \dfrac{\log{m}}{\log(2n-1)} > 1 - \varepsilon$$
or equivalently:
$$ \dfrac{\log(\frac{m}{2n - 1})}{\log(2n - 1)} > -\varepsilon$$
or equivalently:
$$ \dfrac{m}{2n - 1} > (2n - 1)^{-\varepsilon} $$
It is thus sufficient to find a $3AP$-less set with density $\delta := \dfrac{m}{n}$ satisfying:
$$ \delta \geq 2(2n - 1)^{-\varepsilon} $$
Such large $3AP$-less sets exist. See Theorem 3.1 (Behrend, 1946) of the following paper:
http://wiki.math.toronto.edu/TorontoMathWiki/images/2/2d/Expo_paper.pdf
QED #ReductionToAKnownProblem
Best Answer
Try a countable union of sets (such as Cantor sets) whose Hausdorff dimension tends to 1.