I think the answer is, yes, the graph can be connected.
By definition, if the graph G is not connected, then we can find disjoint open sets A and B, such that G is contained in A union B. In particular, that implies no point in G can be contained in the boundary of A.
So to make the graph connected, it's enough to construct an additive function f whose graph intersects the boundary of every open set. To do so, consider a basis H for R as a vector space over Q. This set has cardinality of the reals. Now note that the set of all open sets in R^2 also has cardinality of the reals. (http://en.wikipedia.org/wiki/Cardinality_of_the_continuum)
Put these two sets (basis H, all open sets) in 1-1 correspondence, so for each h in H, we have an open set O(h). We can always find a nonzero rational q, and a real y such that (qh,y) is in the boundary of O(h). Define f(qh)=y. Doing this for all elements of H will determine a unique additive function f on the reals.
The graph of f, by construction, is connected since it intersects the boundary of every open set in R^2. One small note: I guess it's conceivable this construction could give you a continuous f, but you can rule that out by "holding out" a couple basis elements and setting the f values appropriately.