I think the number of holes is not greater than two times the number of sets $C_i$.
Note that the sets $C_i$ can be extended to $E_i$ in a such way that all holes will be convex (see page 8 in paper of Rom Pinchasi, On the perimeter of $k$ pairwise disjoint convex bodies contained in a convex set in the plane. http://www2.math.technion.ac.il/~room/ps_files/perim_k_convex.pdf). Lets think about $C_i$ as about already extended sets.
Each side of a hole is formed by one of the $C_i$-s. Lets call two sets $C_i$ and $C_j$ neighbors, if they form two adjacent sides of some hole. Note that:
- To each hole correspond at least 3 neighbor-pairs - since each hole has at least 3 sides.
- To each neighbor-pair $C_i,C_j$ correspond at most two holes - since all such holes must have a side co-linear with the segment in which the boundaries of $C_i$ and $C_j$ intersect.
Therefore, the number of holes is at most 2/3 the number of neighbor-pairs.
The neighborhood sets form a planar graph, so the number of "neighboring" is (at most) approximately $3n$.
Hence the number of holes is (at most) approximately $2n$.
The lower bound gives the tiling shown on the figure
