Existence of a large leaf in a foliation of the ball Consider a smooth $k$-dimensional foliation of the unit ball $B$ of $\mathbb{R}^n$, all of whose leaves are diffeomorphic to $k$-disks. 
Question: Is there a leaf whose $k$-volume is at least $\omega_k$?
Here $\omega_k$ denotes the volume of the unit ball of $\mathbb{R}^k$.
Particular cases. If $k=1$, the leaf through the origin (in this case a curve) has the desired property. If $k=n-1$, the leaf which divides the ball into two regions of equal volume has the desired property, by a relative isoperimetric inequality. 
A possible argument. Replace each leaf $F$ by a $k$-submanifold which minimizes the $k$-volume among those with boundary $F\cap \partial B$. Among the minimal submanifolds obtained in this way, consider one which passes through the origin: its $k$-volume is at least $\omega_k$ by the monotonicity formula, and the $k$-volume of the leaf with the same boundary is even larger. 
I know how to make this argument rigorous when the foliation is close enough to the foliation by parallel affine subspaces (but a reference or a more elementary proof also for this perturbative case would be very useful). In general, besides for possible singularities of the minimizers (which should not disturb), the problem I see is how to guarantee that at least one of them passes through the origin.  
 A: I think one might be able to prove this  without using the heavy duty machinery from the geometric measure theory that Anton mentioned by modifying the proof of the main result of the following paper of Gromov "Isoperimetry of Waists and Concentration of Maps".
He proves a stronger version of your statement but for the sphere instead a disk:
for any continuous map $f: \mathbb S^n\to\mathbb R^{k}$ there is a fiber $F$ such that $vol(U_\epsilon(F))\ge vol (U_\epsilon(\mathbb S^{n-k}))$ for any $\epsilon>0$. 
This of course immediately gives some lower bound on the volume of the maximal leaf in your case because the double of a disk is bilipschitz to a sphere but I think it's likely that one can  get a sharp bound too.
There is a very readable explanation of Gromov's proof by one of his students posted on the arxiv. That's what I looked at as reading Gromov's papers can be tough.
The construction is roughly as follows. He looks at convex polyhedral partitions of $\mathbb S^n$ into $2^N$  (with $N\to\infty$) subsets of equal volume. Given a map  $f: \mathbb S^n\to\mathbb R^{k}$ he defines a section of a certain vector bundle over the space of partitions and checks that this bundle has a nonzero top Stiefel-Whitney class so that this section must have a zero. by construction a zero of the section gives a fiber of $f$ passing through the center of mass of every convex set in the partition. The argument is very similar to (and is in fact a generalization of) the proof of the Borsuk-Ulam theorem.
Furthermore one can make sure that the convex sets are $\delta(N)$ close to being k-dimensional with $\delta(N)\to 0$ as $N\to\infty$. One then passes to the limit to get a "partition" into convex sets of dimension $\le k$ and the fiber in question is the one that passes through the center of mass of every convex set  with respect to the limit of the normalized volume measure. The key geometric part is to prove that for every $k$-dimensional convex set $C$ in the limit and the normalized limit measure $\mu$ on it one has that $$\mu(B_\epsilon(p)\cap C)\ge \frac {vol (U_\epsilon(\mathbb S^{n-k}))}{vol (\mathbb S^n)}$$ where $p$ is the center of mass of $C$.
Since the elements of the partitions along the sequence had equal volume this yields the result.
The proof of the inequality in the displayed formula is a convexity argument similar to the proof of the Bishop-Gromov volume comparison.
I haven't checked the details but I think the whole construction can be adapted to a disk instead of a sphere by using convex partitions of the disk obtained by coning off the elements of the partitions of the sphere to give  a sharp bound for the disk. This may even be discussed somewhere in Gromov's paper.
