Is the $F_4$ lattice (i.e. 4dimensional bodycentered hypercubic lattice, spanned by the simple roots of $F_4$) bipartite? And if so, what is a good explicit partition of the vertices?

No such labeling can exist because this lattice contains triangles such as $\lbrace(0,0,0,0), \phantom.(2,0,0,0), \phantom.(1,1,1,1)\rbrace$. This is predictable from the root diagram, which contains the $A_2$ root system. [Note that to identify this "bodycentered hypercubic lattice" with a root lattice we must use the norm $\langle x,x\rangle = \frac12(x_1^2+x_2^2+x_3^2+x_4^2)$ so that the minimal vectors satisfy $\langle r,r \rangle = 2$.] Indeed the nearestneighbor graph of this lattice is not even $3$colorable, because it has $4$cliques such as the regular tetrahedron $$ T = \lbrace(0,0,0,0), \phantom.(2,0,0,0), \phantom.(1,1,1,1), \phantom.(1,1,1,1)\rbrace. $$ This, too, is predictable from the root diagrams once we know that the $F_4$ lattice is the same as the $D_4$ lattice (the $F_4$ roots are vectors of norm $2$ and $4$ in $D_4$): the root diagram of $D_4$ contains $A_3$. In general a root lattice has $n$dimensional simplices spanned by roots if and only if it contains $A_n$. There does exist a $4$coloring: the vectors of norm $4$ (the long $F_4$ roots) span a sublattice $L_0$ of index $4$ isometric with $2^{1/2} D_4$, and we can assign each of its cosets a different color. In fact this $4$coloring is unique up to color permutation. To prove this, first observe that each of the vertices of $T$ must have a different color. The same is true for the tetrahedron obtained from $T$ by replacing $(2,0,0,0)$ by $(0,2,0,0)$, so these two minimal vectors must have the same color. They differ by a long root of $F_4$, and all the long roots are equivalent under lattice isometries. Hence any two vectors that differ by a long root have the same color, whence each coset of $L_0$ must have constant color, QED. 

