I think that this follows from Bourbaki's Éléments de Mathématique. Groupes et algèbres de Lie, Chapitre VIII, §10, Theorem 1 (see below) applied to the adjoint representation. Alas, I cannot provide the google books link because the book that Google Books claims to be this one, is actually Algèbre commutative, Chapitres 5 à 7! (And the "Feedback" link does not allow me to point this out, since in their arrogance, Google does not even allow for the possibility of such an error!)
Théorème 1. --- Soient $V$ un espace vectoriel de dimension finie, $\mathfrak{g}$ une sous-algèbre de Lie réductive dans $\mathfrak{gl}(V)$, $\mathfrak{q}$ une sous-algèbre de Lie de $\mathfrak{g}$ et $\Phi$ la forme bilinéaire $(x,y) \mapsto \mathrm{Tr}(xy)$ sur $\mathfrak{g} \times \mathfrak{g}$. On suppose que l'orthogonal $\mathfrak{n}$ de $\mathfrak{q}$ par rapport à $\Phi$ est une sous-algèbre de Lie de $\mathfrak{g}$ composée d'endomorphismes nilpotents de $V$. Alors, $\mathfrak{q}$ est une sous-algèbre parabolique de $\mathfrak{g}$.
And here's a possible translation:
Theorem 1. --- Let $V$ be a finite-dimensional vector space, $\mathfrak{g}$ a reductive Lie subalgebra of $\mathfrak{gl}(V)$, $\mathfrak{q}$ a Lie subalgebra of $\mathfrak{g}$ and $\Phi$ the bilinear form $(x,y) \mapsto \mathrm{Tr}(xy)$ on $\mathfrak{g} \times \mathfrak{g}$. If the orthogonal complement $\mathfrak{n}$ of $\mathfrak{q}$ relative to $\Phi$ is a Lie subalgebra of $\mathfrak{g}$ consisting of nilpotent endomorphisms of $V$, then $\mathfrak{q}$ is a parabolic subalgebra of $\mathfrak{g}$.
Edit
As Fran points out in the comments below, my original translation was incorrect and had $\mathfrak{n}$ nilpotent instead of consisting of nilpotent endomorphisms. Happily, for the case of the adjoint representation, one has Engel's theorem, which says that the the two notions agree. Happily, for the case of the adjoint representation, one has Engel's theorem, which says that the the two notions agree.