Suppose that $K$ is the figure eight knot in the three-sphere $S^3$. Let $n(K)$ be a small open neighbourhood of $K$. Let $X = S^3 - n(K)$. You can find a Heegaard diagram for $X$ as follows.
Let $N = N(K)$ be a closed neighbourhood of $K$, slightly larger than $n(K)$. Draw both in a knot diagram of $K$. Add to $N$ a small neighbourhood $P$ of an "unknotting arc" (also called a "tunnel"). So $V = N \cup P$ is a handlebody, and $V' = N \cup P - n(K)$ is a compression body. Note that $W = S^3 - \mathrm{interior}(V)$ is a handlebody as well.
So $(V', W)$ is a genus two Heegaard splitting of $X$. It is an exercise to give a diagram for this splitting. (Hint: The diagram has only three curves, not four. To find it, start by drawing a knot diagram for $K$. Now add the unknotting arc $\alpha$ along the core of $A$.) Any non-longitudinal Dehn filling of $X$ gives a rational homology three-sphere. Most of these are not integral homology spheres. The filling slope gives the fourth curve in the desired Heegaard diagram.
Note that something similar works for all two-bridge knots and links (and tunnel number one knots, in general). Doing this for torus knots will yield many Seifert fibered examples - the examples above are mostly hyperbolic.