Let $A$ be a commutative ring with unit. A non-degenerate symmetric bilinear form $\phi$ on a finitely generated projective $A$-module $P$ is called metabolic if there is a direct summand $L$ of $P$ such that the adjoint map of $\phi$ $$ \phi' \colon P \to P^\ast = Hom(P,A) $$ restricts to $0$ on $L$ and $L = L^\perp$, where $L^\perp = \ker(P \stackrel{\phi'}{\to} P^\ast \to L^\ast)$. In different terms $L^\perp$ is the submodule of $P$ consisting of elements $p$ such that $\phi(p,L)=\{0\}$. In this case $(P,\phi)$ is isometric to $L\oplus L^\ast$ with bilinear form of the type $((l,\lambda),(k,\kappa)) \mapsto \kappa(l) + \lambda(k) + \gamma(\lambda,\kappa)$, where $\gamma$ is a symmetric (possibly degenerate) bilinear form on $L^\ast$.
Now $(P,\phi)$ is called stably metabolic if there is a matebolicmetabolic module $(N,\nu)$ such that the orthogonal sum $(P,\phi)\perp(N,\nu)$, which is just $(P \oplus N,\phi\oplus \nu)$, is metabolic. The stably metabolic modules are presicelyprecisely the ones that become $0$ in the Witt-group $W(A)$ of symmetric bilinear forms over $A$.
My question: When is a stably metabolic module in fact metabolic? Are there conditions I can put on $A$ to make this true?
(As usual $2$ being a unit in $A$ will probably make everything much easier but I'm trying to avoid this.)