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.

The field is unimportant. The given condition implies that $\operatorname{tr}(APDP^{-1}) = \operatorname{tr}(P^{-1}APD) = 0$ for every invertible matrix $P$ and every diagonal matrix $D$. Since the set of all diagonalisable matrices spans the whole matrix space (there is actually an even better result, namely, every square matrix is the sum of at most three diagonalisable matrices), the previous observation in turn implies that $\operatorname{tr}(AB)=0$ for every matrix $B$. Hence $A=0$.