A conjecture regarding the eigenvalues of real symmetric matrices
The result can be proven using the same arguments as in the proof that the eigenvalues of symmetric matrices are real.
Let $x\ne0$ solve $Mx=\lambda Dx$ for some possibly complex $\lambda$.
If $Dx=0$, then it holds $Mx=0$, and $\lambda$ can be arbitrary. This corresponds to an infinite generalized eigenvalue of the matrix pencil $(M,D)$. Under your assumption, that $M$ and $D$ do not have common null space, then this case cannot happen.
Now let $Dx\ne0$. Since $D$ is diagonal with non-negative entries, it follows $x^HDx\ne0$. Then $$ \lambda x^HDx = x^HMx = (Mx)^Hx= (\lambda Dx)^Hx=\bar\lambda x^HDx. $$ This implies $\lambda=\bar\lambda$, and $\lambda$ is a real number.
Note, that this argument also works if $D$ is assumed to be symmetric positive semidefinite, as then $Dx\ne0$ implies $x^HDx\ne0$.