Here's a proof for odd dimension $n\ge 3$ that the Lie algebras are isomorphic only when the quadratic forms are equivalent up to rescaling (I assume $K$ has characteristic zero and fix an algebraically closed extension $C$).
Let $f:\mathfrak{so}(\phi)\to \mathfrak{so}(\psi)$ be a $K$-defined isomorphism. We can assume that both $\mathfrak{so}(\phi)$ and $\mathfrak{so}(\psi)$ are $K$-defined subalgebras of $\mathfrak{sl}_n$ preserving a nondegenerate quadratic form on the $n$-dimensional space. Then both are $C$-conjugate to $\mathfrak{so}(n)$. Since I assume $n$ odd, over $C$, all automorphisms of $\mathfrak{so}(n)$ are inner. It follows that $f$ can be realized by a conjugation, namely there exists $A\in\mathrm{GL}_n(C)$ satisfying: $f(g)A=Ag$ for all $g\in \mathfrak{so}(\phi)$. The set of $A$ satisfying this condition is a $K$-defined linear subspace on which the determinant map does not vanish; hence it contains a $K$-point with nonzero determinant. That is, $A$ can be found in $GL_n(K)$. Hence $\mathfrak{so}(\psi)$ preserves the $K$-defined quadratic form $x\mapsto \phi'(x):=\phi(A^{-1}x)$. Since the set of $K$-defined invariant forms is 1-dimensional (because the standard representation of $\mathfrak{so}(\psi)$ is absolutely irreducible), it follows that $\phi'$ and $\psi$ are collinear.
A similar argument still works when the outer automorphisms of $\mathfrak{so}(\phi)$ can be realized over $K$, that is, setting $G=\mathrm{Aut}(\mathfrak{so}(\phi))$, when the natural homomorphism $G_K\to (G/G^0)(C)$ is onto. (Recall that the latter group has order 1 for odd $n$, order 2 for $n\ge 4$ except $n=8$, and order 6 for $n=8$.) I don't know examples for which it's not the case but I haven't yet thought about it.