$\langle Tu\;|\;u\rangle=0,\;\forall u\in E \Longrightarrow T=0$?

Suppose $\newcommand{\ip}[2]{\langle #1\mid #2\rangle}\ip{Tu}{u}=0$ for every $u$. Then, given $u$ and $v$ and any scalar $a$ (assuming semilinearity in the first variable, with the opposite convention the proof is essentially the same): $$ 0=\ip{T(u+av)}{u+av}= a\ip{Tu}{v}+\bar{a}\ip{Tv}{u} $$ In particular, for $a=i$, we get $\ip{Tu}{v}=\ip{Tv}{u}$ and, for $a=1$, $\ip{Tu}{v}=-\ip{Tv}{u}$.

Therefore $\ip{Tu}{v}=0$, for every $u$ and $v$, in particular for $v=Tu$. Hence $Tu=0$.

Note that this can fail on real Hilbert spaces; the easiest example is $$ T\colon\mathbb{R}^2\to\mathbb{R}^2 \qquad T\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix}-y\\x\end{bmatrix} $$

A sesquilinear form $s(x,y)$ on a complex vector space can be recovered from the associated quadratic form $q(x)=s(x,x)$ through the polarization identity $$ s(x,y) = \frac{1}{4}\sum_{n=0}^{3}i^n q(x+i^n y). $$ So, if the quadratic form is $0$, then so is the sesquilinear form.

In your case, $s(x,y)=\langle Tx,y\rangle$ and $q(x)=\langle Tx,x\rangle$. So it is in fact the case that $T=0$ if $\langle Tx,x\rangle=0$ for all $x$, but this relies on having a complex space. Real spaces do not behave in the same way.