So, since $Q$ is Hermitian, it can be written as $Q=U\Lambda U^\text{H}$, where $\Lambda$ contains the eigenvalues (which are negative) and U is unitary ($U^HU=I$). Then
\begin{equation}
(x,Qx) = x^H U\Lambda U^H x = (U^Hx) \Lambda (U^Hx)
\end{equation}
We are done actually: since all eigenvalues are negative we have
\begin{equation}
(U^Hx) \Lambda (U^Hx) \leq (U^Hx) A (U^Hx)
\end{equation}
where $A=\text{diag}(a,a,\ldots,a)=aI$. So
\begin{equation}
(x,Qx) \leq a (U^Hx) (U^Hx) = a ||U^Hx|| = a||x||=a(x,x)
\end{equation}
Here I also used the fact that, since U is unitary it does not affect the length of vectors.