Can you cover a genus a billion hyperbolic surface with 15 balls?

Your conjecture is false. Every nonorientable closed connected surface of negative Euler characteristic, admits a hyperbolic metric such that the surface is covered by 3 embedded disks. Hence, for each $p\ge 2$, there is a closed connected orientable hyperbolic genus $p$ surface covered by 6 embedded disks. (One can likely reduce this number to 3 as well.) Here is a construction. Start with a regular hyperbolic $2n$-gon $P$ with $n\ge 4$ and the vertex angle $2\pi/n$. (I assume that $P$ is closed and 2-dimensional.) Let $R$ denote the radius of the inscribed circle in $P$ and $a$ the common length of the sides of $P$.

Denote its vertices $z_kw_k$, indexed cyclically by $k\in {\mathbb Z}_n$. Then there is a collection of hyperbolic isometries $g_k$ sending (the oriented segment) $z_kw_k$ to $z_{k+1}w_{k+1}$ and, respectively, $h_k$ sending $w_kz_{k+1}$ to $w_{k+1}z_{k+2}$ and mapping the interior of $P$ to the exterior of $P$. Note that under the identification via these isometries, there are two equivalence classes of vertices of $P$. I assume, you are familiar with Poincare's Fundamental Domain Theorem: It shows that the isometries $g_1,...,g_n, h_1,...,h_n$ generate a discrete torsion-free isometry group of the hyperbolic plane with fundamental domain $P$ and quotient surface $X$ homeomorphic to $N_p$ (the nonorientable surface of genus $p$), $p=n-1$. Just in case, my favorite reference is

Beardon, Alan F., The geometry of discrete groups, Graduate Texts in Mathematics, 91. New York - Heidelberg - Berlin: Springer-Verlag. XII, 337 p. DM 108.00; $ 44.60 (1983). ZBL0528.30001.

Next, take the (open) inscribed hyperbolic disk $B$ in $P$ (of the radius $R$), centered at the center $o$ of $P$. Furthermore, take closed hyperbolic disks $A_k, C_k, k=1,...,n$ centered at, respectively, the vertices $w_k, z_k$ of $P$ and whose radii equal $a/2$. All these $2k+1$ disks completely cover $P$ and the $A$-disks are pairwise disjoint and $C$-disks are pairwise disjoint. Intersect these disks with $P$. (I will retain the names $A_k, C_k$ for the intersections. A picture would be great here but would take too much effort to draw.)

The unions $A_1\cup...\cup A_n$, $C_1\cup...\cup C_n$ project to two isometrically embedded closed disks $A, C$ in $X$; I will denote the projection of $B$ again by $B$; it is again isometrically embedded since I chose $B$ to be open. Then $A\cup B\cup C=X$. Thus, we got two closed and one open disk in $X$ covering $X$. In order to get three open disks with the same properties, expand slightly $A$ and $C$ and take the interiors of the expanded disks.

Question: Is there a function $V(n,m)$ such that for every compact hyperbolic $n$-manifold $X$, $n\ge 3$, covered by $m$ embedded hyperbolic metric disks, the volume of $M$ is $\le V(n,m)$?

Consider a regular hyperbolic $4n+2$-gon with vertex angle of genus $3$. I claim we can glue $3$ copies of this polygon together to form an oriented hyperbolic surface of genus $n$, without gluing any polygon to itself.

This will form a decomomposition of the surface with $4n+2$ vertices, $6n+3$ edges, and $3$ faces, for an Euler characteristic of $4n+2 + 3 - (6n + 3 ) = 2-2n$.

I will describe it by first describing the vertices, then the edges, then the cyclic ordering of the edges at each vertex, and finally the faces.

The vertices are indexed by the numbers $0$ to $4n+1$, mod $4n+2$. There are edges from vertex $i$ to vertices $i+1, i-1, i+2n+1$. Thus each vertex has three edges.

The three edges leaving $i$ towards vertices $i+1, i-1, i+2n+1$ are in clockwise order.

One face has boundary vertices $0,1,2,3,\dots, 4n+1$ in order (clockwise around the face. Another has boundary the vertices $$ 2n+1 , 2n , 4n+1, 4n, 2n-1, 2n-2, 4n-1, 4n-2, 2n-3,2n-4, \dots, 1,0$$ in order (clockwise around the face).

The last one has boundary the vertices $$ 4n+1, 2n, 2n-1, 4n, 4n-1 , 2n-2, 2n-3, \dots, 2n+1,0$$ in order (clockwise around the face).

Each edge meets two of the three faces and each vertex meets all three faces. So indeed the polygons glue to a smooth hyperbolic surface.

Now take a disc whose center is the center of each polygon, whose radius is at least the distance from the center to a vertex but less than the distance from a center to a vertex plus half the length of an edge). The disc will contain the polygon, hence three will cover the surface, but each disc will only extend a short distance from each polygon into the next and thus will not intersect itself.