No, the null set of such a function need not have non-empty interior. In fact, for any closed subset $C\subset \mathbb R$, there exists $u\in C^\infty(\mathbb R)$ such that $u^{-1}(0) = C$:
Proof: We can write the complement of $C$ as a countable, disjoint union of open intervals $\mathbb R \setminus C = \bigcup_n I_n$. Now choose cut-off functions $\psi_n\ge 0$ such that $\psi_n(x) > 0$ iff $x\in I_n$.
Next, choose coefficients $c_n>0$ such that $c_n |\psi_n |_{C^n(\mathbb R)}\le 2^{-n}$ for $n=1, 2, \dots$ Then the function $u(x) = \sum_{n=1}^\infty c_n\psi_n(x)$ satisfies $u(x)>0$ iff $x\in \bigcup_n I_n$ and $u(x) = 0$ otherwise. That is to say $u^{-1}(0) = C$.
Furthermore for any $k\in \mathbb N$, we have $|u|_{C^k(\mathbb R)} \le \sum_{n\ge1} c_n|\psi_n|_{C^k(\mathbb R)} \le \sum_{n< k} c_n|\psi_n|_{C^k(\mathbb R)} + \sum_{n\ge k} 2^{-n} < \infty$ i.e. $u\in C^\infty(\mathbb R^n)$. $\square$
Now you can take $C$ a fat Cantor set in $[0,1]$ and find a $C^\infty$ function $u$ satisfying $u^{-1}(0) = C$. This $u$ is then in $W^{1,p}_0((0,1))$, but $C$ is nowhere dense.