EDIT: I explain below why a flat hypersurface implies ruled, but the converse is not true. See Robert Bryant's answer for a more precise explanation.
It is a theorem that any 2-dimensional open surface in $R^3$ that has zero Gauss curvature everywhere (and therefore is a flat 2-dimensional Riemannian manifold) is a ruled surface (i.e., is foliated by straight line segments). Such surfaces apparently are called developable surfaces.
I have not bothered to check the details but I am betting that everything generalizes in a straightforward manner to an $n$-dimensional hypersurface in $R^{n+1}$. Such a hypersurface has a flat Riemannian metric if and only if the sectional curvature vanishes on all tangent 2-planes. But the sectional curvature of any tangent 2-plane is equal to the determinant of the restriction of the second fundamental form at that point to the 2-plane. Therefore, the hypersurface is flat if and only if the rank of the second fundamental form is at most 1. The proof that a 2-dimensional surface is ruled probably extends to show that any flat hypersurface in higher dimensions is foliated by codimension 2 planes. So these could be called developable hypersurfaces. In other words, a hypersurface is flat if and only if it consists of a 1-parameter family of codimension 2 planes.
ADDED: I have confirmed that the above is right. In particular, any neighborhood of a point where the second fundamental form is not identically zero can be foliated uniquely into codimension 2 planes. This is a nice and rather easy exercise in the local geometry of a hypersurface in Euclidean space using the Frobenius theorem.