For $k>1$ ($k=2$ in particular for reason) and $\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_{n}}=1$ (it is $a_{n} > 0$ as well, but that is not crucial) we compare these two series:

$\sum\limits_{n=1}^{+\infty}(-1)^n ((\frac{a_{n+1}}{a_{n}})^k-1)$

$\sum\limits_{n=1}^{+\infty}(-1)^n (\frac{a_{n+1}}{a_{n}}-1)$

Can we claim anything about the convergence of the second knowing the convergence of the first?

You can take if needed be the restrictions: $a_{n+1} > a_{n}$ and even more $a_{n} \in \mathbb{N}$

Of course, we can say that there should be a general requirement that $a_{n+1}$ does not differ in any critical sense from $a_{n}$. For example we may ask that we these two converge

$\sum\limits_{n=1}^{+\infty}(-1)^n ((\frac{a_{2n+2}}{a_{2n}})^k-1)$

$\sum\limits_{n=1}^{+\infty}(-1)^n ((\frac{a_{2n+1}}{a_{2n-1}})^k-1)$

or something similar that makes certain that we are dealing with convergent series even when split. (Do not take this split convergence as the requirement. This is just an example of possible restrictions. Very good if this one is working. Even better if none is needed.)

This seems plausible, and for any nice example out there it works just fine, but I do not see any direct way of proving it. I suspect that there must be something very wrong with the series if this is not working.

(A normal monotone requirement for alternating series should play no role in the proof as neither of the series needs to be monotone. Only $a_{n}$ can be monotone.)

Let me display the reasoning behind this question. Set $k=2$ in the question as this allows reducing the exponent on its own and it allows to have:

$\sum\limits_{n=1}^{+\infty}(-1)^n ((\frac{a_{n+1}}{a_{n}})^2-1)$ $=\sum\limits_{n=1}^{+\infty}(-1)^n (\frac{a_{n+1}}{a_{n}}-1)(\frac{a_{n+1}}{a_{n}}+1)$

Now if we take the condition that for any $N$

$|\sum\limits_{n=1}^{N}(-1)^n (\frac{a_{n+1}}{a_{n}}+1)| < M$

the result might follow if we have something like Dirichlet test just in reverse and that would not require monotone condition or some combination of conditions.

Can this last help to establish maybe the opposite result under some conditions, that if

$\sum\limits_{n=1}^{+\infty}(-1)^n (\frac{a_{n+1}}{a_{n}}-1)$

converges then

$\sum\limits_{n=1}^{+\infty}(-1)^n ((\frac{a_{n+1}}{a_{n}})^2-1)$

converges?