I think you are basically there:

Let $U$ be the unitary operation you defined, $U \phi(x) U^\dagger \equiv -\phi(x)$. This is a definition. We then assume (or check) that this operation is a symmetry of the Lagrangian. 

We now **assume** that the ground state $|\Omega\rangle$ is also invariant under this operation, $U |\Omega\rangle = |\Omega\rangle$ (there is in principle a phase factor, but we can take it to be $1$ by redefining $U$). 

With this assumption, 
\begin{align}
  \langle \Omega|\phi(x)|\Omega\rangle
      &= \langle \Omega| U \phi(x) U^\dagger |\Omega\rangle \\
      &= \langle \Omega| (-\phi(x)) |\Omega\rangle \\
      &= -\langle \Omega|\phi(x)|\Omega\rangle,
\end{align}
which then implies $\langle \Omega|\phi(x)|\Omega\rangle = 0$.