Are there uncountably many disjoint uncountable real null sets?

As $C$ is topologically the only compact zero-dimensional metric space without isolated points, $C \simeq C \times C$, so every Cantor set is a union of continuum many disjoint homeomorphic copies of itself ($\{c\} \times C, c \in C$). So such an uncountable union can be a null set (but need not be, beware!).

First do it for $C=\{0,1\}^{\mathbb{N}}$. For every $b=(b_n)_n\in \{0,1\}^{\mathbb{N}}$ let $C_b=\{(a_n)\ \mid\ a_{2n} = b_n \textrm{ for all } n\}$. Clearly $(C_b)_b$ form a partition of $C$ into compact subset of measure $0$.

Now $[0,1]$ is measure equivalent to $C$ under the map $a\mapsto s(a) = \sum a_n/2^{n+1}$. We can eliminate the countable subset of "undetermination" of $s^{-1}$ (the binary fractions) by considering the images $s(C_b)$, where $b$ is a sequence that is not eventually constant. We get in this way an uncountable family $(s(C_b))_b$ of compact disjoint subsets of $[0,1]$ of measure $0$. The complement of their union in $[0,1]$ is again uncountable and of measure $0$.

Note that for $b$ not an eventually constant sequence we can describe $s(C_b)$ to be the set of those real numbers in $[0,1]$ such that their binary expansion $\sum a_n/2^{n+1}$ has $a_{2n}= b_n$ ( prescribed digits at odd positions after the binary dot).

While at it, it should be clear how to partition each $s(C_b)$ ($b$ not eventually constant) into an uncountable family of Cantor subsets.

The answer is yes: Let $C_r\subset \mathbb R^2$ for $r\in\mathbb R$ be defined as $$C_r=\{(r,c)\mid c\in \mathcal C\}$$(where $\mathcal C$ is the Cantor set). Then, consider a continuous surjection from $\mathbb R\to\mathbb R^2$. The inverse images of each $C_r$ are clearly uncountable and disjoint, and because the function is continuous, they are still zero sets.