No. Let $$u=\begin{bmatrix}1\\\1\end{bmatrix}$$ $$A=\begin{bmatrix}1 & 0 \\\ 0 & 2\end{bmatrix}$$ $$B=\begin{bmatrix}2 & 0 \\\ 0 & 1\end{bmatrix}$$ Then $$u'Au=u'Bu=3$$
No. Let $$u=\begin{bmatrix}1\\\1\end{bmatrix}$$ $$A=\begin{bmatrix}1 & 0 \\\ 0 & 2\end{bmatrix}$$ $$B=\begin{bmatrix}2 & 0 \\\ 0 & 1\end{bmatrix}$$ Then $$u'Au=u'Bu=3$$