Must a matrix of which all conjugates have zero diagonal be zero?

I think it is true: Suppose $A$ is nonzero. Then we find nonzero $v,w\in \mathbb R^n$ with $Av=w$. If $w$ and $v$ are linearly dependent, extend $v$ to a basis, then $A$ written in that basis will have a nonzero entry in the diagonal. If they are independent, then so are $v$ and $v+w$. Extending $\{v,v+w\}$ to a basis will then yield a nonzero diagonal element in $A$ written in this basis.

An alternative, coordinate-based answer: Conjugate with the mutually inverse matrices $S^\pm_{ij}=\delta_{ij}\pm\delta_{i\alpha}\delta_{j\beta}$ for arbitrary $\alpha\ne\beta$ (that is, the matrices where $\pm1$ is added to the unit matrix in some off-diagonal entry) and consider the $\alpha$-th diagonal element of the result:

$$ \left.\sum_{jk}\left(\delta_{ij}+\delta_{i\alpha}\delta_{j\beta}\right)A_{jk}\left(\delta_{kl}-\delta_{k\alpha}\delta_{l\beta}\right)\right|_{i=l=\alpha}=\left.\sum_{jk}\left(\delta_{ij}+\delta_{j\beta}\right)A_{jk}\delta_{kl}\right|_{i=l=\alpha}=A_{\alpha\alpha}+A_{\beta\alpha}\;. $$

Clearly $A_{\alpha\alpha}=0$ (from conjugating with the unit matrix), and since $\alpha\ne\beta$ were arbitrary, it follows that all entries of $A$ are zero.