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 hyperbolic genus $p$ surface covered by 6 embedded disks. (One can almost surely reduce this number to 3 as well.) Here is a construction. Start with a regular hyperbolic $2n$-gon $P$ with $n\ge 3$ and the vertex angle $2\pi/n$. (I assume that $P$ is closed and 2-dimensional.)

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 $p$). 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 $P$, 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 the half-edge lengths of $P$. 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.

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 $\le V(n,m)$?