The answer to the question, as asked, is "no".  

For, suppose that $B$ is a three-ball. So, $B$ is a genus zero handlebody.  Let $\alpha$ be a knotted arc properly embedded in $B$.  So the fundamental group of $B - \alpha$ is not free.  Let $A$ be the frontier of a regular neighbourhood of $\alpha$.  So $\alpha$ is an annulus, properly embedded in $B$.  Cutting $B$ along $A$ and taking closures yields a three-ball ("inside $A$") and a knot complement ("outside $A$").  

In general, we can take $H$ to be an handlebody of any genus and carry out the above procedure inside of a ball inside of $H$.  Or we can use arcs $\alpha$ that do not lie in a three-ball in $H$ - there is no simple classification of knotted arcs in handlebodies. 

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However, the annulus $A$ is compressible, and perhaps you want to rule that out?  If we add the assumption that $A$ is incompressible, then the answer to the question becomes "yes".

For, suppose that $H$ is a handlebody.  Suppose that $A$ is a properly embedded incompressible annulus.  Suppose that $D$ is a collection of essential disks, properly embedded in $H$, cutting $H$ into a collection of three-balls.  We properly isotope $A$ to minimise the number of arcs of $A \cap D$.  

We now cut $H$ along $D$.  The components of $A$ (in the resulting three-balls) are all disks (as otherwise $A$ is compressible).  

We now cut $H - D$ along $A$.  By Alexander's lemma (and after taking closures) the result is a collection of three-balls.  We call these the "pieces" of $H - (A \cup D)$.  The remains of $\partial H$, $A$, and $D$ on the boundaries of the pieces are called "patches".  So all patches are planar surfaces.  In fact, all $D$- and all $A$-patches are disks (by the incompressibility of $A$ and the minimality of $A \cap D$).  

We now reassemble the pieces of $H - (A \cup D)$, gluing only along the $D$-patches.  So, we are gluing three-balls along disks: this gives a union of handlebodies.