First of all there are isometries of symmetric spaces (of noncompact type) which have positive displacement, but no invariant geodesics. This means that if you replace surface groups by cyclic groups, there are no (in general) equivariant harmonic maps. Using this example, one can build an example of a surface group representation without an equivariant harmonic map. For instance, let $G=SL(2,R)\times SL(2,R)$. Take $\rho=(\rho_1,\rho_2)$, where $\rho_1$ is the uniformizing representation of $\Sigma_g$ and $\rho_2$ is a representation whose image is a cyclic parabolic group. One can check that this representation is $P$-Anosov. (Here $P$ is the product of Borel subgroup in the first $SL(2,R)$-factor and of the entire second $SL(2,R)$-factor.) On the other hand, there is no equivariant harmonic map for $\rho$ since $\rho_2$ does not have one.
However, if your $P=B$ is Borel, then any $B$-Anosov representation is reductive and, hence, admits an equivariant harmonic map.
If you just want to have a continuous extension of an equivariant continuous map to $G/K$, then (for any $P$-Anosov representation) such extension exists, provided you use the appropriate topology of "flag-convergence" on
G/K \sqcup G/P.$$
This is contained e.g. in my paper with Bernhard Leeb and Joan Porti "A Morse Lemma for quasigeodesics in symmetric spaces and euclidean buildings". For this you do not need surface group representations, any $P$-Anosov representation of any hyperbolic group would be OK.
Lastly, the Furstenberg boundary of $G/K$ is $G/B$, while for general parabolic subgroup $P$, the quotient $G/P$ is a partial flag-manifold of $G$.