You've received two good answers, but I'll elaborate a bit.
Usually equidistribution on the torus (or more general, compact groups) wrt the Haar measure is achieved by computing the Weyl sums and showing that there are some cancellations.

The question you are referring to is studied in the area of so-called "sparse equidistribution" (although it is more of sparse density if you would like).

The problem with the harmonic-analytic approach is by summing (or integrating) over very sparse part of your period. It is usually not stright forward to bound such exponential sums.
For example, Vinogradov proved for example that $ \{p_{n}x\} $ is equidistributed mod 1 for all irrational x, to bound the Weyl sums, he used sieves with what is called now Vinogradov sums, and a result about the odd Goldbach conjecture.

Now if you are interested in a metric result (i.e. a.e. x), then it is a very classical result that for every increasing unbounded sequence \$ {a_{n}}$ and for a.e. x, one have that ${a_{n}x}$ is equi. mod 1, this is done by taking the Weyl sum, computing its L^2 norm, and then sub. limit and integration by the DCT.

Now the question if such a result follows for every x is very subtle, and not always amenable to harmonic-analytic approach, and the current state of the art actually lies in the ergodic approaches.

If you have a sequence which is contained inside a geometric progression, then there exists x's for which $\{a_{n}x\}$ is not equi. more generally, for $\{q^{n}\}$ say, you can find x's whose orbit closure is with any Hausdorff dimension you want (the reason here that as a dynamical system, this is isomorphic to Bernoulli shift on $q$ letters).
More generally, a result due to Boshernitzan says that if you have a lacunary sequence (the limit of the ratios of consecutive elements is larger than 1), then the Hausdorff dimension of the set of exceptional x's (such that $\{a_{n}x\}$ is not dense/equidistributed) is 1.
On the contrary, Boshernitzan shown that if the sequence is non-lacunary (the ratio tends to 1, you should think about it as having sub-exp. growth), then the Hausdorff dimension of the set for which $\{a_{n}x\}$ is equi. is 1.
There were even some old results due to Erdos from the 1950's about it (he talked about convolution of Bernoulli measures, which can be interpreted in this sense as well).

A very peculiar discovery by Furstenberg (67) shown that if you have a non-lacunary semigroup, say $\{2^{n}3^{m}\}$ then for every irrational x you get that $\{2^{n}3^{m}x\}$ is dense mod 1 (certainly not equidistributed).
This result is very interesting, because you have density for every x. Moreover, recently, Bourgain-Lindenstrauss-P. Michel and Venkatesh proved an effective version of that theorem (meaning that you fix some epsilon, you can estimate how far you need to go in-order to find an element which is epsilon-close to an integer).
An even recent work (by myself, still preprint), generalizing the Bourgain-Lindenstrauss paper, and I shown that sets like $\{2^{n}3^{3^{m}}3^{3^{k^2}}x\}$ are dense for every x.

About Fibonacci sequences, it follows from my work (based on other work of D. Meiri whith Yuval Peres and Elon Lindnstrauss), that you can prove density of sequences such as $\{2^{n}3^{3^{m}}F_{k}x\}$ for every irrational x.
For the general Fibonachi sequences, you are basically in the lacunary case, which Boshernitzan already covered in certain sense.