Since people have addressed (2-4), I'll address (1). There is indeed a fast argument.
Namely, you have a flat family of curves $X\to \operatorname{Spec}(\mathcal{O}_{K, p})$. You are interested in comparing the geometric genus of the generic fiber $X_K$ to that of the special fiber $X_{\mathbb{F}_q}$. Note that by constancy of Euler characteristic in flat families, $$1-p_a({X_K})=\chi(\mathcal{O}_{X_K})=\chi(\mathcal{O}_{X_{\mathbb{F}_q}})=1-p_a(X_{\mathbb{F}_q}).$$where $p_a$ denotes arithmetic genus. So the arithmetic genera of the fibers are equal. Since you've required that the fibers $X_K$ and $X_{\mathbb{F}_q}$ are smooth, their geometric genera equal their arithmetic genera, which completes the argument.
ADDED (12/19/2012): I'd also like to add a comment about the Hodge decomposition for curves, to complement Christian's excellent answer. Namely, one can see that the Hodge-theoretic predictions one would make from the situation over $\mathbb{C}$ are true for smooth projective curves over any base. A rigorous version of this is given (with a sketch proof) as Theorem 2 in these notes of mine.
