It is not: for example, the piece-wise constant function f: [0,1] -> [0,1] which starts at f(0) = 0, stays constant until suddenly f(1/2) = 1, until f(3/4) = 0, until f(7/8) = 1, etc. is Riemann integrable.
"Continuous almost everywhere" means that the set of its discontinuities has Lebesgue measure 0. Many infinite sets have Lebesgue measure 0, including all countable sets.
"iff it is bounded and has countable discontinuities"?
Or, are there some uncountable sets which also have Lebesgue measure 0?
The indicator function of the Cantor set is Riemann integrable. Like you said, though, the Dirichlet function (which is the indicator function of the rationals) is not Riemann integrable.
The reason is because the Dirchlet function is discontinuous everywhere on [0,1], so the set of discontinuities has measure 1. The Cantor function is discontinuous only on the Cantor set.
Likewise, the indicator function of a "fat Cantor set" (a way of constructing a Cantor-like set w/ positive measure) is not Riemann integrable: https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%9...