I got an answer from MathOverflow. A counterexample can be constructed using The Devil's staircase. If one replaces $\hat{I}$ by $I$, than the Cantor function is Holder continuous surjection and it is easy to construct a full measure set K (the set where the function is locally constant = the union of middle third intervals = the set of numbers whose base 3 expansion contains at least one 1) that is mapped to a zero measure set (the set of numbers with base 2 expansion having only finite nonzero coordinates).
To understand that $K$ has measure $1$ look at $I\setminus K$. It is a set of numbers without a single $1$ in base-3 expansion. I will show that this set has measure zero due to standard ergodic theory. Base $3$ expansion of a number $y$ can be contructed by assigning a sequence of digits $\{\sigma_k\}$ from $\{0,1,2\}^\mathbb{N}$ to the trajectory of $y$ under the action of $E_3 (x) = 3x (\mod 1)$.
If $E_3^{k} (x) < 1/3$, put $\sigma_k = 0$, if $1/3 < E_3^{k} (x) < 2/3$, put $\sigma_k = 1$, and so on.
As $E_k$ is ergodic with respect to Lebesgue measure for every $k\geq 2$, for Lebesgue almost every $x$ the trajectory of $x$ under the action of $x \to 3x (\mod 1)$ visits every open set (including middle third).
To understand that $f(K)$ has zero measure one can use a similar argument -- for Lebesgue almost every $x$ the trajectory of $x$ under the action of $x \to 2x (\mod 1)$ visits every open set (including second half) infinitely many times.
To handle the case of Hilbert cube one can take a composition of the Cantor function and a projection.
In fact, there are other similar examples here
