Number of minima in a ribbon disk?
A handle decomposition of a ribbon disk gives a band presentation of the knot. If I understand how this goes correctly, the band presentation is an unlink along with attached bands between them such that
performing embedded arc surgery along these bands (which carry framing information for the surgery) gives the knot, and
if the unknots are thought of as vertices and the bands as edges, the graph forms a tree (since loops in the graph would correspond to a maxima, which don't exist).
The number of components in the unlink is the number of minima, and so your $I(K)$ is the minimal number of such components over all band presentations of the ribbon knot $K$.
For example, here is a band presentation of a square knot:
By the way, there appear to be some moves that can go between any two band presentations for the same ribbon disk of a ribbon knot:
(See at least Figure 3 of https://arxiv.org/abs/1804.09169 or the paper they cite: Swenton, Frank J., On a calculus for 2-knots and surfaces in 4-space, J. Knot Theory Ramifications 10, No. 8, 1133-1141 (2001). ZBL1001.57044.). I wonder what additional moves you might need to go between any two ribbon disks.
Anyway, there is a nice way to generate ribbon knots with fundamental groups that surject onto a group finitely generated by the conjugates of some element, as you are probably aware:
Johnson, Dennis, Homomorphs of knot groups, Proc. Am. Math. Soc. 78, 135-138 (1980). ZBL0435.57003.
Edit/warning: I tried fixing what follows, but I'm not sure it's salvageable. I had thought that if you looked at Johnson's construction carefully, you could get the result that each $I(K)$ was an upper bound for $\operatorname{rank}(\pi_1(S^3-K))$ when $K$ is a ribbon knot. However, I failed to consider that there might be a band presentation with fewer minima by knotting up the bands, which is a source of additional rank!
Each generator is assigned an unknot, and each band sum introduces a single additional relation depending on which unknots the band passes through. If we start with some knot $K$, then there is a ribbon knot $K'$ such that there is a surjection $\pi_1(S^3-K')\to \pi_1(S^3-K)$, which implies that $I(K')\geq \operatorname{rank}(\pi_1(S^3-K))$, where the rank is the minimal number of generators over all presentations of the finitely generated group $\pi_1(S^3-K)$.
The following paper implies that there are knot groups of arbitrary rank:
Weidmann, Richard, On the rank of amalgamated products and product knot groups, Math. Ann. 312, No. 4, 761-771 (1998). ZBL0926.20019.
In particular, the $n$-fold connect sum of any nontrivial knot $K$ has $\operatorname{rank}(\pi_1(S^3-\mathop{\#}_nK))\geq n+1$. Thus, by following Johnson's construction one gets an infinite family of ribbon knots $K_1,K_2,\dots$ with $I(K_n)\geq n$.
(If there were a way to modify Johnson's construction to generate prime knots, then one could get an infinite family of prime ribbon knots with arbitrary rank.)
(This answer used to claim that $I(K)\geq \operatorname{rank}\pi_1(S^3-K)$ for a ribbon knot, but now I'm not so sure about that.)