This is not really an answer, more of an extended comment, but it was getting too long. Besides, if I am mistaken then it is easier to downvote or correct something posted as an answer. $\newcommand{\norm}[1]{\Vert#1\Vert} \newcommand{\Real}{{\bf R}} \newcommand{\ptp}{\hat{\otimes}}$
I think that in the original question, if one drops the symmetry requirement, then the answer is negative in general. Here's my reasoning.
Claim #1: if X is a Banach space then the bilinear maps $\Psi:X\times X \to \Real$ which are bounded by the bi-sublinear functional $\beta(x,y) = \norm{x}\ \norm{y}$ are in one-to-one correspondence with the linear functionals on the projective tensor product $X\ptp X$ that have norm $\leq 1$.
(This is somehow the defining property of the proj. t.p. -- or at least, it characterizes it up to isometric isomorphism.)
Claim #2: there exists a Banach space $X$ and a closed subspace $E$ such that the natural restriction map $(X\ptp X)^* \to (E\ptp E)^*$ is not surjective.
(I admit that I can't recall the details of an example, but I think taking $X$ to be $L^1$ and $E$ to be the span of the Rademachers would work.)
Now, take some $\psi\in(E\ptp E)^\ast$ which is not in the image of the restriction map, and which satisfies $\norm{\psi}\leq 1$. Let $\Psi$ be the corresponding bilinear functional on $E$. Clearly $\Psi(x,y) \leq \norm{x}\norm{y}=\beta(x,y)$. But any bilinear extension of $\Psi$ to a $\beta$-dominated functional on $X\times X$ would correspond to an element of $(X\ptp X)^*$ which extends $\psi$, and our choice of $E$, $X$ and $\psi$ ensures this can't happen.