It is independent of ZFC. As you mention, CH implies the answer is yes. A different axiom, MA+$\neg$CH, implies the answer is no.

A little more precisely, there is a cardinal number denoted $\mathrm{cov}(\mathcal M)$, one of the so-called ``small cardinals'', that is defined to be the smallest number of meager sets needed to cover $[0,1]$. MA implies this cardinal is equal to the continuum, and if that is the case (and CH fails) then the answer to your question is no, because a partition of $[0,1]$ as you describe is also a covering of $[0,1]$ with meager sets.

But this is not the end of the story. There is a difference between covering and partitioning, and in general it seems harder to partition $[0,1]$ into copies of the Cantor space than to cover it with copies of the Cantor space. I have a paper with Arnie Miller (link) where we look at this question (and some similar ones), and show that it is consistent with the continuum being very large that for every $\kappa < \mathfrak{c}$ there is a partition of $[0,1]$ into exactly $\kappa$ copies of the Cantor space.

**EDIT:** I just realized that Taras Banakh has asked a very similar question here. The statement of his question actually contains an answer to your question, and you can learn a lot more about this question by looking at his question, the comments under it, and the answer I posted there.

disjointunion of $\aleph_1$ Borel sets with empty interior (a theorem of Hausdorff), but this is not necessarily so for $\aleph_2 \leq \kappa < \mathfrak{c}$. The Cantor set can be partitioned into $\mathfrak{c}$ sets, but if you try to put $\kappa$ of them into a partition of $[0,1]$, you're left needing to do something with the complement of those $\kappa$ sets. It might be really ugly (badlynon-Borel), and you might not be able to cover it with fewer than $\mathfrak{c}$ extra Borel sets. $\endgroup$ – Will Brian Feb 1 '18 at 17:27