Given an undirected, unweighted graph with n vertices and m edges, you can decide whether an orientation with minimum indegree at least 2 exists (and find it, if so) in O(m^3) time. If there are no vertices of degree 3, you can sometimes do this fairly simply in O(m) time.
If v is a vertex of degree 1, the task is impossible. If v has degree 2, both its edges must be out-edges, so we can delete 'v' from our graph for now. We do this operation repeatedly based on the new degrees; it's possible that we end up with new degree-3 vertices, but we can now assume all vertices have degree >= 3.
If all vertices are degree 4 or greater, the orientation must exist, and you can use a standard Eulerian circuit algorithm (available in many standard graph libraries) as described in this post. You may need to modify the graph first: Eulerian circuits only exist if all vertex degrees are even. There are always an even number of odd degree vertices: match them up in pairs in any way, adding an edge for each pair. Orienting the edges on the original graph based on the circuit gives a solution for your problem.
For graphs with degree-3 vertices, you can combine a standard algorithm with a nonstandard one. The standard algorithm is maximum-cardinality matching for general graphs, usually the 'Blossom algorithm', which can be found in standard graph libraries. The non-standard (i.e., probably not found in your graph library) algorithm is found in the paper On finding orientations with the fewest number of vertices with small out-degree, which reduces the problem of 'Find an orientation with the fewest outdegree-less-than-2 vertices' to the problem of finding a maximum matching in a graph with O(m) vertices. The algorithm is quite short, and simpler since our graph has minimum degree 3. This solves your problem for fewest-indegrees-less-than-2 after you flip all edge directions.
