No. This is explicitly stated in the paragraph above Theorem 1 of:

[Turner, Edward C, Test words for automorphisms of free groups.
Bull. London Math. Soc. 28 (1996), no. 3, 255--263.][1]

The author refers to Proposition 1.

**EDIT:**

Let's give an explicit example.  Let $F=\langle a,b\rangle$ and let $g=a[b,a]=ab^{-1}a^{-1}ba$.  Clearly $\langle g\rangle$ is a retract of $F$.  But the Whitehead graph of $g$ is two triangles glued along an edge.  This is Whitehead-reduced, so $g$ is not part of a free basis for $F$.


  [1]: http://blms.oxfordjournals.org/cgi/reprint/28/3/255.pdf