Yes, a monotonic function on an interval is Riemann integrable. In fact, one can come up with the step functions naively.
Let $P_n$ denote the partition $\{a, a + \frac{b-a}{n}, a + 2\frac{b-a}{n}, \dots, b\} =: \{x_0, \dots, x_{n}\}$. Then it's easy to see that $x_k - x_{k-1} = \frac{b-a}{n}$. Suppose wlog that $f$ is monotonic decreasing (increasing is more or less the same, or do the same reasoning on $-f$). Then if $M_k$ and $m_k$ denote the maximum and minimum on the $k$th part of the partition, then $M_k = f(x_{k-1})$ and $m_k = f(x_k)$.
Thus
1. (lower sum) $\displaystyle L(f,P_n) = \sum m_k(x_k - x_{k-1}) = \sum f(x_k)\frac{b-a}{n}$ and
2. (upper sum) $\displaystyle U(f,P_n) = \sum M_k(x_k - x_{k-1}) = \sum f(x_{k-1})\frac{b-a}{n}$
So then $\begin{align} U(f,P_n) - L(f,P_n) &= \frac{b-a}{n}\left( \sum_{k = 1}^n f(x_{k-1}) - f(x_k)\right) = \\ &= \frac{b-a}{n} (f(x_0) - f(x_n)) = \frac{b-a}{n}(f(a) - f(b)) \\ &\to 0 \text{ as } (n \to \infty) \end{align}$
It's not so hard to show that refinements of these partitions do not change the result.
You also ask whether a monotonic function on $[a,b]$ is continuous a.e. It is, in the sense that there are only a countable number of discontinuities. In particular, you might know that a sum of positive numbers is finite only if it has a countable number of nonzero terms (an easy way to see this - if there are uncountably many, then you can show there are uncountably many rationals as follows: consider the intervals implied by the sum, where each summand gives another interval. Each interval contains a rational.) The function is bounded on $[a,b]$ and monotonic, and so the sum of the 'jumps' at the points of discontinuity is finite. Thus there are at most countably many points of discontinuity.