$\newcommand{\from}{\colon}\newcommand{\Aut}{\rm{Aut}}\newcommand{\bdy}{\partial}$Here is a fairly hands-on proof. I'll use the following notation: $H = H_g$, $S = S_g = \bdy H$, and $F = F_g$.  I'll write $g(S) = g(H) = g$.  Also, $i \from S \to H$ is the inclusion map. 

> Suppose that $f_* \from \pi_1(S) \to F = \pi_1(H)$ is an epimorphism.  Then there is a homeomorphism $k \from S \to S$ so that $i_* \circ k_* = f_*$.

**Fact number one**: since $S$ and $H$ are Eilenberg-MacLane spaces, there is a map $f \from S \to H$ inducing $f_*$.  Fix a *cut system* $D \subset H$: that is, a collection of $g(H)$ disks so that $H - n(D)$ is a three-ball. Homotope $f$ to make it transverse to $D$.  Thus $C = f^{-1}(D)$ is a collection of simple closed curves in $S$. For any $\alpha \subset C$, let $D(\alpha)$ be the disk containing $f(\alpha)$. 

We now construct a graph $G$.  We have a vertex $v(X)$ for every component $X$ of $S - C$.  We have an edge $e(\alpha)$ for every component $\alpha$ of $C$.  We connect one (both) end(s) of $e(\alpha)$ to $v(X)$ if one (both) side(s) of $\alpha$ are contained in $X$.  Note that $G$ is a finite connected graph. 

Recall that the genus $g(S)$ is the number of curves we must cut $S$ along to obtain a connected planar surface.  In similar fashion we define the *genus* $g(G)$ to be the number of edges we must remove from $G$ to obtain a tree.  Define a map $r \from S \to G$ as follows.  All points of $X - n(C)$ are mapped to $v(X)$.  The open annulus $n(\alpha)$ is sent to the edge $e(\alpha)$ by crushing the circle coordinate.  Deduce that $g(G) \leq g(S)$.  We also observe that $r$ induces an epimorphism $r_* \from \pi_1(S) \to \pi_1(G)$, a free group of rank $g(G)$. [This much of the proof implies that $\pi_1(G)$ can only surject free groups of at most half the rank of $\pi_1(S)$.]

Fix a point $x \in H - D$.  We now define a map $h \from G \to H$ as follows.  The map $h$ sends all vertices of $G$ to $x$.  The edge $e(\alpha)$ is sent to any loop based at $x$ and meeting $D$ exactly once, transversely, at a point of $D(\alpha)$.  (I am ignoring a small issue about orientations here.)  We deduce that $f_* = h_* \circ r_*$.  Thus $h_* \from \pi_1(G) \to F$ is an epimorphism.  **Fact number two**: since free groups are Hopfian, $g(G) = g(S)$ and $h_*$ is an isomorphism.

We now construct a handlebody $H'$, homotopy equivalent to $G$.  Take $S$, thicken to get $S \times [0,1]$, and attach two-handles along the curves $C \times \{0\}$.  We next attach three-handles to the sphere components of the lower boundary.  We identify the groups $\pi_1(H')$ and $\pi_1(G)$.  Let $i' \from S \to H'$ be the inclusion map.  

**Fact number three**: using Nielsen's theorem (giving generators of $\Aut(F)$), there is a homeomorphism $h' \from H' \to H$ inducing $h_*$.  Consider the homeomorphism $k = (\bdy i)^{-1} \circ \bdy h' \circ \bdy i'$, where all domains and ranges are restricted to be $S$.  Chasing the diagram of spaces shows that $i_* \circ k_* = f_*$, and we are done.