You are correct this function is not Riemann integrable on $[0,1]$. Your reasoning is (almost, see edit) correct, but another quick way to check is to note that its set of discontinuity points, namely $(0,1]$, does not have Lebesgue measure zero.
It is, however, Lebesgue integrable. We have $f(x) = x\chi_{\mathbb{Q} \cap [0,1]}(x) -x\chi_{[0,1]\setminus\mathbb{Q}}(x)$ which is a sum of bounded measurable functions. Since $[0,1]$ has finite measure bounded and measurable implies integrable.
Edit: Your reasoning is almost correct. Showing $U-L>0$ always does not imply the function is not Riemann integrable. You need to show $U-L > \epsilon$ always, for some $\epsilon >0$.
Also, another quick way to see that $f$ is Lebesgue integrable is that $f(x)=-x$ almost everywhere. Since the Lebesgue measure is complete, this implies $f$ is integrable if and only if $g(x):= -x$ is integrable, which of course it is.