I think the best way to see the signature of these quadratic forms is by using the formula from "A Classification Theorem for Albert Algebras" by R. Parimala, R. Sridharan, and Maneesh L. Thakur, Trans. AMS 350 #3, March 1998.

All forms of $F_4$ arise from Albert algebras.  Over $R$, these are 27-dimensional algebras, whose unital automorphisms form groups of type $F_4$.  They are classified, over fields of characteristic neither $2$ nor $3$, by cohomological invariants $f_3$ and $f_5$.  These cohomological invariants determine 3-fold and 5-fold Pfister forms, $\phi_3$ and $\phi_5$ respectively.

The formula of P-S-T (above), or maybe originally due to Serre, is that for an Albert algebra $A$ over $k$,
$$Q_A \perp \phi_3 \cong <2,2,2> \perp \phi_5.$$

Now there are only two Pfister forms over $R$ for $\phi_3$ and $\phi_5$.  The signature of $\phi_3$ is either $(8,0)$ or $(4,4)$.  Similarly, the signature of $\phi_5$ is either $(32,0)$ or $(16,16)$.  The signature of $<2,2,2>$ is $(3,0)$.  Hence the possibilities for the signature $(p,n)$ of $Q_A$ are:
$$(p,n) + (8,0) = (3,0) + (32,0),$$
$$(p,n) + (8,0) = (3,0) + (16,16),$$
$$(p,n) + (4,4) = (3,0) + (32,0),$$
$$(p,n) + (4,4) = (3,0) + (16,16).$$

Only three cases are possible:  $(p,n) = (27,0)$ or $(p,n) = (11,16)$ or $(p,n) = (15,12)$.

As $F_4$ acts on the orthogonal complement of the identity, and the identity has positive norm, the possible signatures for the 26-dimensional rep of $F_4$ are:
$$(26,0), (10,16), (14,12).$$