Witt's theorem states that
If ($V,Q)$ and $(V',Q')$ are isomorphic and nondegenerate, every injective metric morphism $$s: U \rightarrow V' $$ of a subvector space U of V can be extended to a metric isomorphism of $V$ onto $V'$.
But I think that there is a counterexample. Let $k=\mathbb R$ and $V=V'=k^2$. Let $Q$ be a standard one, that is $Q(x_1,x_2)=x_1^2+x_2^2$ and $Q'(x_1,x_2)=2x_1x_2$. Each of the induced bilinear form is a standard dot product and $(x,y)_{Q'}=x_1y_2+x_2y_1$. Let $U=<(1,0)>$ and $s(t,0)=\frac{t}{\sqrt 2}(1,1)$. Then $s$ is a metric morphism. However, let $S$ denote the extension of $s$, $S(0,1)=(x_1,x_2)$.
First, $Q'(x_1,x_2)=2x_1x_2=1 \Rightarrow x_1x_2=\frac{1}{2}$.
Second, $((x_1,x_2),(\frac{1}{\sqrt 2},\frac{1}{\sqrt 2}))_{Q'}=((1,0),(0,1))=0 \Rightarrow x_2=-x_1$.
This is a contradiction. So, why this is 'not' a counterexample?
$Q$ and $Q'$ are manifestly non-isomorphic since they don't even have the same range. ($Q'$ has negative numbers in its range and $Q$ does not.)