If three nonzero real matrices mutually anticommute, then at least one of them has a negative off-diagonal element

This isn't true, at least when the matrices are allowed to be singular. Let $$ D=\pmatrix{1&0\\ 0&-1}, \ S=\pmatrix{0&1\\ 1&0}. $$ Then $DS$ and $SD$ are nonzero but $DS+SD=0$. The following set of matrices now serves as a counterexample to your statement: $$ X_1=\pmatrix{D\\ &D\\ &&0}, \ X_2=\pmatrix{S\\ &0\\ &&D}, \ X_3=\pmatrix{0\\ &S\\ &&S}. $$