Let $Q=0$. Then we have for all $x,y\in E$: \begin{eqnarray*} 0 &=& Q(x+y) - Q(x-y)\\ &=& \langle Ax+Ay, x+y\rangle - \langle Ax-Ay, x-y\rangle\\ &=& 2\langle Ax,y\rangle + 2\langle Ay,x\rangle\\ &=& 2\langle Ax,y\rangle + 2\langle y,A^*x\rangle\\ &=& 2\langle Ax,y\rangle + 2\langle A^*x,y\rangle\\ &=& 2\langle (A+A^*)x,y\rangle, \end{eqnarray*} where we used the fact that $E$ is a real Hilbert-space. Since $x,y$ were arbitrary, this implies that $A+A^* = 0$, i.e. $A^* = -A$. Now assume that $A$ is skew-symmetric, i.e. $A^*=-A$. Then we have for all $x\in E$: $$ \langle Ax,x\rangle = \langle x,A^*x\rangle = \langle A^*x,x\rangle = -\langle Ax,x\rangle, $$ and thus, $Q(x) = \langle Ax,x\rangle = 0$ for all $x\in E$.