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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.

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 $D$ 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.

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.

However, in the above example the annulus $A$ is compressible. 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 $D$ 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.

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.

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.

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.

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 $D$ 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.

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 $A$").

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.

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.