Can there be a manifold with uncountable genus? Can they be compact? Just some random thoughts I had after coming across this paper on Riemann Surfaces of Infinite Genus: [http://www.math.ubc.ca/~feldman/papers/allriem.pdf]
I'm reasonable convinced that $\mathbb{R}^n$ cannot have uncountably many regions (of non-zero measure) removed, but I don't quite see that this ruins the potential for manifolds which can be sometimes a lot more whacky than $\mathbb{R}^n$.
Do the Homotopy Groups or Holonomy Groups or the Cohomology Groups still make any sense in this context?  I think they'd have infinitely many generators, which would get strange. 
 A: There's nothing particularly strange about homotopy groups or homology groups having a countable infinity of generators (unless the manifold is compact, as said in the comment of @LordSharktheUnknown). 
For example the ladder surface is the boundary of a regular neighborhood in $\mathbb R^3 = \mathbb R^2 \times \mathbb R$ of 
$$\bigl((\mathbb R \times \{0,1\}) \cup (\mathbb Z \times [0,1])\bigr)  \times \{0\}
$$ 
The ladder surface has countably generated $\pi_1$ and $H_1$, but neither is finitely generated. Perhaps an even simpler example is $\mathbb R^2$ minus the radius $1/3$ balls centered at the points of $\mathbb Z \times \{0\}$.
Here's a general fact along these lines. Connected smooth manifolds are usually required by definition to be paracompact (this is unnecessary for Riemann surfaces, which are proved to be paracompact by Rado's Theorem). One can then use a partition of unity argument to prove that there is a "good open covering" by open sets homeomorphic to balls which is locally finite and such that any finite subset of the covering intersects in either the empty set or a subset homeomorphic to a ball. By applying combination theorems (e.g. Van Kampen's theorem for $\pi_1$ and the Mayer Vietoris theorem for $H_n$, perhaps coupled with direct limit arguments) one can then conclude that $\pi_1$ and each $H_n$ have countable generating sets (any of them could, nonetheless, still be finitely generated). By applying somewhat deeper combination theorems one can also prove that the higher homotopy groups $\pi_n$ have countable generating sets.
