Implications of orthonormal wavefunctions
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.