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 $\langle \Omega|\phi(x)|\Omega\rangle = 0$.

I think you are basically there:

Let $\hat{U}$ be the unitary operation you defined, $\hat{U} \hat{\phi}(x) \hat{U}^\dagger \equiv -\hat{\phi}(x)$. 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, $\hat{U} |\Omega\rangle = |\Omega\rangle$ (there is in principle a phase factor, but we can take it to be $1$ by redefining $\hat{U}$).

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

Edit:

This is my attempt at showing that a $\hat{U}$ as above exists. I'm denoting operators in Fock space with a hat, $\hat{U}$, while unhatted operators act in the single-particle Hilbert space.

For a non-interacting theory, the field operator can be expanded as \begin{align} \hat{\phi}(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}[\hat{a}_{\mathbf{p}}e^{i\mathbf{p}\cdot{\mathbf{x}}} + \hat{a}_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot{\mathbf{x}}}], \end{align} where $\hat{a}_{\mathbf{p}}$ denotes the annihilation operator for the mode with momentum $\mathbf{p}$.

A unitary transformation which preserves particle number acts on the annihilation operators as $\hat{U} \hat{a}_{\mathbf{p}} \hat{U}^\dagger = \int dp^\prime U(\mathbf{p};\mathbf{p}^\prime) \hat{a}_{\mathbf{p}^\prime}$, where $U(\mathbf{p};\mathbf{p}^\prime)$ are the matrix elements of the single-particle representation of the unitary operation. In particular, for $U(\mathbf{p};\mathbf{p}^\prime) \equiv -\delta(\mathbf{p} - \mathbf{p^\prime})$, we have $\hat{U} \hat{a}_{\mathbf{p}} \hat{U}^\dagger = - \hat{a}_{\mathbf{p}}$. A similar calculation holds for $\hat{a}^\dagger_{\mathbf{p}}$. Using this in the expression for $\hat{\phi}$ we obtain that $\hat{U} \hat{\phi}(x) \hat{U}^\dagger \equiv -\hat{\phi}(x)$, as required.

Then there is the question of how does this generalize to an interacting theory, which I don't know.