Question: Is there a function $f:[0,1]\rightarrow R$ such that for all $B\subset [0,1]$ with the Lebesgue measure $\mu(B)>0$, $f$ is unbounded on $B$?

Answer: 
Yes. Let $A$ be a Vitali set, that is, a set of representatives of $\mathbb R/\mathbb Q$, and let $m:\mathbb Q\to\mathbb N$ be an enumeration of the rationals. Define $f(\alpha+q)=m(q)$ for all $\alpha\in A, q\in\mathbb Q$ (restrict to $\alpha+q\in[0,1]$ if you want a function defined on $[0,1]$). There is no set $B$ with $\mu(B)>0$ on which $f$ is bounded, because each such set would be a subset of $\bigcup\_{n=1}^N f^{-1}(n)$ for some $N$, and all measurable subsets of a non-measurable set $f^{-1}(n)$ necessarily has measure $0$ (see [The subset of non-measurable set](https://math.stackexchange.com/questions/88449/the-subset-of-non-measurable-set) ).
It seems so. For the sake of simplicity I consider the unit circle $\mathbb T=\{z\in\mathbb C:|z|=1\}$ instead of the unit segmet $I=[0,1]$. Now we determine a standard partition of the set $\mathbb T$ into countably many non-measurable subsets. At first, define an equivalence relation $\sim$ on $\mathbb T$, putting $e^{\varphi i}= e^{\psi i}$ iff $(\varphi-\psi)/\pi$ is rational. Let $A\subset T$ be a set intersecting with each class of the equivalense $\sim$ exactly in one point. Then $T=\bigcup\{A\_q:q\in [0;2)\cap\mathbb Q\}$, where $A\_q=Ae^{q\pi i}$ is a partition of $\mathbb T$ into countably many congruent subsets. Let $e:[0;2)\cap\mathbb Q\to\mathbb N$ be an arbitrary bijection. Now, at last, define a function $f:\mathbb T\to \mathbb R$ putting $f(z)=e(q)$ iff $z\in A\_q$. Suppose that there exists a measurable subset $B$ of $\mathbb T$ such that $\mu(B)>0$ and $f|B$ is bounded. Then there exists a finite subset $F$ of $\mathbb Q$ such that $B\subset\bigcup\{A\_q:q\in F\}=C$. But since $\mathbb T$ contains countably infinitely many disjoint copies of $C$, we see that $\mu(B)=0$, a contradiction.