I'll give the details of the abstract proof, just to emphasize that proving uncountability is much easier than proving perfectness. This is not the same as Nishant's answer, but is more or less the same as user6976's comment.
The set of cube-free words is a subshift, i.e. shift-invariant and closed in the Cantor space $\{0,1\}^{\mathbb{N}}$. Clearly it has no periodic points, since an infinite word like $uuu...$ contains in particular the word $u^3$.
Now a basic fact:
Theorem. A nonempty subshift that has no periodic points is uncountable.
Proof. Take a minimal subsystem $Y$ (a nonempty subsystem that has no proper nonempty closed subsystems), it's a classical theorem that one exists. It is easy to show that minimality is equivalent to "every forward orbit enters every open set". From this characterization it is easy to see that a minimal system is either a single periodic orbit or is perfect, namely if $\{y\} \subset Y$ is open, then $\sigma(y)$ eventually enters it, and $y$ is periodic. Now, a perfect subset of Cantor space is uncountable, so $Y$ is uncountable, and it is contained in our original subshift so that one is also uncountable. Square.
In particular the argument is not specific to cube-free words.
Corollary. Let $\alpha > 1$ and $A$ a finite alphabet. If there exists an $\alpha$-power free infinite word $x$ over alphabet $A$, then there exist uncountably many such words.
Note that this does not tell you that the set of such words is perfect. The cube-free words do form a perfect set as explained by Currie, but the proof is much more intricate.