Ostrowski's theorem give the answer for valuations, but is there a complete classification of (at least separated) topologies on Q (compatible with the field operations, obviously)?
According to this link, there are as many field topologies on $\mathbb{Q}$ as there are subsets of $\mathbb{R}$, so I doubt there is a classification. A reference seems to be Wieslaw's book "topological fields" (it doesn't seem to be on google books, unfortunately).
PS: note that there is only one nonHausdorff field (or ring) topology on a field (the two open sets one) since the closure of 0 is an ideal, and the others are completely regular, as any Hausdorff group topology. Also there is only continuum many metrizable topologies on $\mathbb{Q}$, so most of the topologies referred to in the above link are quite pathological, non first countable for instance.
PPS: searching "number of field topologies" in MathSciNet returns the following references
Podewski, KlausPeter The number of field topologies on countable fields. Proc. Amer. Math. Soc. 39 (1973), 33–38.
Kiltinen, John O. On the number of field topologies on an infinite field. Proc. Amer. Math. Soc. 40 (1973), 30–36.
and they are both freely accessible (thanks AMS!) here and here.
Concerning the proof for a countable field $K$, Podewski manages to define a continuum $\mathcal{G}$ of (metrizable) field topologies on $K$ such that the suprema of any two distinct susbsets of $\mathcal{G}$ are distinct field topologies.
The details are somewhat complicated, but the idea is quite natural. Suffice it to say that the metrizable field topologies on $K$ are parameterized  via fundamental sequences of neighbourhoods of $0$  by chains $G$ in a partially ordered set $P$ of "conditions" (is it forcing in disguise?), which specify for a finite number of elements of $K$ wether they belong or not to the $n$th neighborhood in the sequence.
Strangely, for uncountable fields $K$ the proof is easier, using valuations instead of chains of conditions, and the fact that the transcendence degree of $K$ over the prime subfield is the cardinality of $K$.

$\begingroup$ It's nice to have an answer of course; moreover, EOM is pretty reputable and the answer seems highly believable, but: is there a simple proof for the first sentence? $\endgroup$ – Todd Trimble♦ Sep 6 '12 at 3:46

1$\begingroup$ @Todd: I was also dissatisfied, so I added a PPS. The proof is not that long (around 10 pages), but not very simple either, at least for me. Perhaps it would look simpler to a model theorist. $\endgroup$ – BS. Sep 6 '12 at 12:03

2$\begingroup$ I object to the suggestion "non metrizable" $\Longrightarrow$ "quite pathalogical". $\endgroup$ – Gerald Edgar Sep 6 '12 at 12:23

1$\begingroup$ @Todd Trimble: I am confused by your response to Gerald Edgar (indeed there was a similar statement as a comment on the question but it got deleted). What you say seems to contradict the content of the PS in the answer ("only one nonHausdorff"). $\endgroup$ – user9072 Sep 6 '12 at 13:39

1$\begingroup$ Yes, as pointed out twice earlier, by BS and quid. Yours makes three times. :) As for your last statement, might "pathological" be in the eye of the beholder? Or do you have something more specific in mind to back up your claim? I'd be genuinely interested. $\endgroup$ – Todd Trimble♦ Sep 6 '12 at 14:41