The following seems to be implied by most of the direct comments to OP's question, but I prefer to voice it, loud and clear:

>> The answer is **yes**, we can find such integers $m = m(\alpha, \varepsilon)$ and $n = n(\alpha, \varepsilon)> 0$, for almost every real number $\alpha > 1$, in the sense of Lebesgue measure.

This is so, because the sequence $(\alpha^n)_n$ is then [equidistributed modulo $1$][1] in the sense of H. Weyl, see "Uniform distribution of sequences" by  L. Kuipers et H. Niederreiter (Wiley, 1974).

Equidistribution modulo $1$ implies density of the set 
$\{\alpha^n - \lfloor \alpha^n \rfloor : n \in \mathbb{N}\} \subset \lbrack 0, 1\rbrack$, which in turn implies that the sequence $(\alpha^n - \lfloor\alpha^n\rfloor)_n$ of fractional parts accumulates on $0$.

As already observed above, for the golden number $\alpha = \frac{1 + \sqrt{5}}{2}$,  the sequence $(\alpha^n)_n$ is not equidistributed modulo $1$ since $\alpha^n + (\frac{1 - \sqrt{5}}{2})^n$ is an integer for every $n$, but fractional parts accumulate on $0$. A similar line of reasoning applies to [Pisot-Vijayaraghavan numbers][2] (this is Robert Israel's comment, see also Noam D. Elkies'comment for [Salem numbers][3]).

It is not known whether $((\frac{3}{2})^n)_n$ is equidistributed modulo $1$ (See Gerry Myerson's comment for references). It is also not known whether $(e^n)_n$ or $(\pi^n)_n$ is equidistributed modulo $1$.
 Actually, no explicit example $\alpha > 1$ is known to be equidistributed modulo $1$. (All my remarks are borrowed from [a lecture on ergodic theory by Pierre de La Harpe][4].)


  [1]: https://en.wikipedia.org/wiki/Equidistributed_sequence
  [2]: https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number
  [3]: https://en.wikipedia.org/wiki/Salem_number
  [4]: https://www.unige.ch/math/folks/delaharpe/vulgarisation/4Therg20mar05.pdf