If you have two orthonormal states $|\psi\rangle$ and $|\phi\rangle$,$$ \langle \psi|\phi\rangle = 0,$$there is no guarantee that the term$$ \langle\psi|\hat{A}|\phi\rangle$$vanishes. All depends on the action of $\hat{A}$ on your state.
As you already noticed, if $|\phi\rangle$ is an eigenstate of $\hat{A}$ with eigenvalue $a_\phi$, the term is obviously zero:$$ \langle\psi|\hat{A}|\phi\rangle = \langle\psi|a_\phi|\phi\rangle =a_\phi \langle\psi|\phi\rangle = 0.$$But this is only a special case, since $|\phi\rangle$ is not necessarily an eigenstate of $\hat{A}$ in general.
One of the simplest situations where this term does not vanish can occour when $|\psi\rangle$ and $|\phi\rangle$ are eigenstates of the harmonic oscillator, that we can denote $|n\rangle$ and $|n'\rangle$. As you probably know, these states are orthonormal:$$ \langle n|n'\rangle = \delta_{n,n'}.$$If you consider the specific case $|\psi\rangle=|n\rangle$ and $|\phi\rangle=|n-1\rangle$, it immediately follows that$$ \langle n|n-1\rangle = 0.$$Now if you take $\hat{A}= \hat{a}^\dagger$, for which you have $\hat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$ ($|n\rangle$ is not an eigenstate of $\hat{a}^\dagger$), you have ($n\geq1$)$$ \langle n|\hat{a}^\dagger|n-1\rangle = \langle n|\sqrt{n}|n\rangle = \sqrt{n} \langle n|n\rangle = \sqrt{n}\delta_{n,n} = \sqrt{n} \neq 0.$$
As I already stated, all depends on the action of $\hat{A}$ on your state.
