You don't mean that the forgetful map contracts any rational component, only those touching less than three nodes. Also, the n-fold fibered power is not singular where several markings coincide at a smooth point. After all, a fibered product of smooth morphisms is smooth. It will, however, disagree with $\overline{M}_{g,n}$ at the locus where three or more markings coincide, so blow-ups will indeed be required.
However, what you wish for in the last paragraph is indeed true. One has to assume that the automorphism group of $C$ is trivial, for otherwise, even the fiber of $\overline{M}_{g,1}$ over $\overline{M}_g$ will be not $C$ but $C/ Aut C$. (Working with moduli stacks rather than moduli spaces would cure this problem.) But assuming this, the fiber of $\overline{M}_{g,n} \to \overline{M}_g$ is indeed the Fulton-MacPherson configuration space. Indeed, that space is described in the original F-Mac paper as a moduli space of stable configurations of distinct smooth points on a fixed curve $C$ with nodal trees of projective lines attached, modulo projective equivalence on the lines. Here "stable" means that the pointed nodal curve has finite automorphism group, that is, each line carries at least three distinguished points. Those are exactly the configurations that you see in your fiber.
As for your earlier question, it is true in some sense, but it would be quite subtle in practice. Fulton-MacPherson explain how to obtain their space from the product $C^n$ by an explicit sequence of blow-ups. One blows up first the "small diagonal" where all the points come together, and then proper transforms of other diagonals. You could try to blow up all the corresponding loci in the fibered power of $\overline{M}_{g,n}$ over $\overline{M}_g$.
But you would have to resolve other loci over the boundary too: if two markings coincide with a node, for example, you want to pull apart the node, glue in an extra line, and draw the two markings on that line. But since there is a 1-parameter family of ways to do this (due to the cross-ratio of the two nodes and the two markings), you want to replace the corresponding points in the fibered power with a $P^1$. Even worse, this would not be a blow-up in the obvious sense. The fibered product of two nodes, in local analytic coordinates, looks like $xy = t = wz$, that is, $xy-wz=0$, the cone on a smooth quadric surface. You can resolve this singularity by blowing up the origin, but then the exceptional divisor is $P^1 \times P^1$, not $P^1$ as you wish. Instead, you should perform one of the "small resolutions" obtained by blowing up one of the Weil divisors $x=w=0$ or $x=z=0$.
Since any birational morphism is the blow-up at some sheaf of ideals, in some abstract sense you are guaranteed that $\overline{M}_{g,n}$ is obtained from the fibered power by blowing up. But that doesn't mean that you can give an explicit list of smooth centers to be blown up in turn. The example with the two markings at the node suggests that this will be a thorny problem.
