It seems that this should be possible. What I had in mind to show that it's possible is by modifying a function that converges to the Dirichlet function so that it's uniform convergent:
Let $\{r_1, r_2, ..., r_n \} $ be an enumeration of all the rationals in $[0,1]$.
define $g_n(x) = \{\frac{1}{n}, x=r_n, $ $0,$ otherwise.$\}$
I'm just not completely sure if this counts as uniformly convergent.
This is for integrable, not necessarily Riemann integrable nor Lebesgue integrable
Thanks!
This is not possible under the stronger assumption that the $ g_n $ are bounded. Indeed, then $ g $ is also bounded by uniform convergence, and by the usual argument, if all of the $ g_n $ are continuous at a point $ x $, then so is $ g $: given $ \epsilon > 0 $, pick $ N $ such that $ \|g - g_N\|_{\infty} < \epsilon/3 $, and pick $ \delta > 0 $ such that for all $ t $ such that $ |x - t| < \delta $, we have $ |g_N(x) - g_N(t)| < \epsilon/3 $. Then, for the same range of $ t $, we have
$$ |g(x) - g(t)| \leq |g(x) - g_N(x)| + |g_N(x) - g_N(t)| + |g(t) - g_N(t)| < \epsilon $$
so that $ g $ is continuous at $ x $. Since the countable union of finite sets is countable, the set of points $ x $ such that at least one of the $ g_n $ is discontinuous at $ x $ is countable; and at all other points, $ g $ is continuous by the above argument. Therefore, the discontinuity set of $ g $ is at most countably infinite; in particular, it has Lebesgue measure zero. By Lebesgue's criterion for Riemann integrability, $ g $ must therefore be Riemann integrable.
If the $ g_n $ are not assumed to be bounded, then one has trivial counterexamples: fixing a function $ f : [0, 1] \to \mathbb R $ with finitely many discontinuities and an infinite integral (for example, $ f(x) = 1/x $ for $ x > 0 $, $ f(x) = 0 $ for $ x = 0 $), we may simply choose $ g_n = f $ for all $ n $.