Let $\mathfrak g$ be a finite dimensional real Lie algebra. If a bilinear map $A:\mathfrak g\times\mathfrak g\to\mathbb R$ vanishes on commuting elements (i.e. $[U,V]=0\implies A(U,V)=0$), is there a linear map $\phi:\mathfrak g\to\mathbb R$ so that $A(U,V)=\phi([U,V])$? If this is not generally true, does it hold for some non-abelian Lie algebras?
I have tried to find a proof or a counterexample, but to no avail. The reason I think this has some hope of being true is this result for linear maps:
Let $E$ and $F$ be real vector spaces and $\alpha:E\to F$ a linear map. If a linear map $A:E\to\mathbb R$ satisfies $\ker\alpha\subset\ker A$, then there is a linear map $\phi:F\to\mathbb R$ so that $A=\phi\circ\alpha$.
Here $\alpha$ plays the role of the commutator. I do not see how to generalize the proof of this linear result to the bilinear realm, especially since the kernel (preimage of zero) of a bilinear map is not generally a vector space.
If two Lie algebras $\mathfrak g$ and $\mathfrak h$ satisfy the desired property, so does their direct sum $\mathfrak g\oplus\mathfrak h$. Abelian Lie algebras satisfy it trivially. Therefore if one shows that all simple Lie algebras satisfy it, the result follows for all reductive Lie algebras. This is such a large class of Lie algebras that I would be satisfied with an answer under the additional assumption that $\mathfrak g$ is simple.
The answer is yes for all Lie algebras.
(It turns out that merely posting a well-thought question at MSE significantly helps answering one's questions. I figured out an answer to my question a couple of hours after posting it.)
First, observe that $A$ vanishing on commuting elements implies that $A$ is skew-symmetric. Writing out $A(U+V,U+V)$ using bilinearity and using $A(U,U)=A(V,V)=A(U+V,U+V)=0$ gives $A(U,V)+A(V,U)=0$ for all $U,V\in\mathfrak g$. Using this observation it is straightforward to see that the desired property indeed holds on $\mathfrak{so}(3)$. However, similar calculation by hand is not feasible in a general Lie algebra.
Consider the canonical bilinear map $\iota:\mathfrak g\times\mathfrak g\to\mathfrak g\otimes\mathfrak g$, given by $\iota(U,V)=U\otimes V$. By the universal property of tensor products, there is a (unique) linear map $\gamma:\mathfrak g\otimes\mathfrak g\to\mathbb R$ so that $A=\gamma\circ\iota$. Let $I$ be the subspace of $\mathfrak g\otimes\mathfrak g$ generated by elements of the form $U\otimes V+V\otimes U$ for $U,V\in\mathfrak g$. The quotient $\mathfrak g\otimes\mathfrak g/I$ is the exterior power $\Lambda^2\mathfrak g$, and denote the quotient map by $q$. We normalize things so that $q(U\otimes V)=U\wedge V$.
Now if $T\in I$, we have $\gamma(T)=0$. (In other words, $\ker q\subset\ker\gamma$.) This is because any such $T$ is a linear combination of elements of the form $U\otimes V+V\otimes U$ and $$ \gamma(U\otimes V+V\otimes U)=\gamma(U\otimes V)+\gamma(V\otimes U)=A(U,V)+A(V,U)=0. $$ Furthermore, define $\beta:\Lambda^2\mathfrak g\to\mathfrak g$ by $\beta(U\wedge V)=[U,V]$. This is a well-defined linear map.
We have $\ker q\subset\ker\gamma$ and $\ker q\subset\ker(\beta\circ q)$, but in fact it is also true that $\ker(\beta\circ q)\subset\ker\gamma$. To see this, observe that $\beta(q(\iota(U,V)))=[U,V]$, so vanishing of $\beta(q(\iota(U,V)))$ implies that of $\gamma(\iota(U,V))=A(U,V)$, and that $\mathfrak g\otimes\mathfrak g$ is spanned by elements of the form $\iota(U,V)$.
Now we are in a position to use the linear result mentioned in the OP, replacing $E\leadsto\mathfrak g\otimes\mathfrak g$, $F\leadsto\mathfrak g$, $\alpha\leadsto\beta\circ q$ and $A\leadsto\gamma$. We have $\ker(\beta\circ q)\subset\ker\gamma$ The result states that there is a linear map $\phi:\mathfrak g\to\mathbb R$ so that $\gamma=\phi\circ\beta\circ q$. Composing this with $\iota$ gives $A=\gamma\circ\iota=\phi\circ\beta\circ q\circ\iota$. This is what we want, since $\phi$ is linear and $$ A(U,V) = \phi(\beta(q(\iota(U,V)))) = \phi(\beta(q(U\otimes V))) = \phi(\beta(U\wedge V)) = \phi([U,V]). $$