I believe that $K$ is an infinitely generated free group. First we find an infinite presentation $K = \langle X \mid R \rangle$, and then we argue that we can pick an infinite subset of $X$ that freely generates.

**Presentation**: Note that we can obtain the Borromean rings from the handlebody of genus two by drilling out the commutator curve and attaching a two-handle to the boundary along the usual separating curve. Since the attaching curve for the two-handle is trivial in the handlebody, we can take the universal cover of the handlebody and then attach infinitely many two-handles, and drill infinitely many lines. After drawing a few pictures, we can organize all of this information as follows.

Let $D^2$ be the Poincare disk model for the hyperbolic plane. Pick hyperbolic isometries $a$ and $b$ that have perpendicular axes and equal (very long) translation lengths. Thus $F_2 = \langle a,b \rangle$ is a free group of rank two. Let $T$ be the regular four-valent tree isometrically embedded in $H^2$, arising as the Cayley graph of $F_2$. Let $v_0$ be the vertex of $T$ corresponding to the identity of $F_2$. The universal cover of the handlebody above is obtained as a three-dimensional neighborhood of $T$, and $F_2$ is the deck group.

We define $X = \{ x_q \}$ where $q$ ranges over the connected components of $H^2 - T$. (These correspond to the drilled lines.) This gives the generating set for $K$ subject to the relations $R = \{ x_p X_q x_r X_s \}$ where the regions $p,q,r,s$ are arranged counter-clockwise about a vertex of $T$. (These correspond to the attached two-handles. Because we started with the commutator curve, the two-handles are attached to the neighborhood of $T$ via a "baseball curve" around each vertex.) Here we use the convention that $X_q = (x_q)^{-1}$. (Also, $p$ is always to the "north-east" of $v$ - we use translates of the picture about $v_0$ to set conventions.) For future use, let $w_v$ be the relation coming from vertex $v$.

**Subset that freely generates**: We recursively color the regions of $H^2 - T$ as follows. (1) All regions start white. (2) In the first step, we pick two regions $p, q$ adjacent to $v_0$. Color both of $p$ and $q$ black. (3) In general, pick a vertex $v$ having exactly two white regions $p, q$ adjacent to $v$. Note that $p$ and $q$ must be adjacent to each other. Color $p$ black and color $q$ grey.

The set of black regions gives a subset $X' \subset X$ of the generators. It is an exercise to prove inductively that $X'$ generates. Next, for a contradiction, suppose that $w$ is a word in the $X'$ that is trivial in $K$. Thus we can write $w$ as a product of conjugates of relations from $R$. Recall that relations are in bijection with the vertices of $T$. Pick one relation $w_v$ appearing in the product so that $v$ is as far as possible from $v_0$. Of the four regions adjacent to $v$, there are two regions $p, q$ that are combinatorially further than the others from $v_0$. Say $p$ and black and $q$ is grey. Then $x_q$ is not freely cancelled in the product of conjugates, a contradiction. QED