As requested by Jukka Kohonen, I'll turn my comment into an answer.  

The answer is in general **no**.   If such an embedding exists, then the red subgraph and blue subgraph are both planar. However, every $n$-vertex planar graph has at most $3n-6$ edges.  So, any $n$-vertex graph with at least $6n-11$ edges is a counterexample.  

The minimum number of colours required for such an embedding is often called *geometric thickness*.  For example, geometric thickness is mentioned on the Wikipedia page for [graph thickness][1], which is the minimum number of planar subgraphs that partition the edge set of a graph (where the embedding is not necessarily the same for each planar subgraph).  


https://en.wikipedia.org/wiki/Thickness_(graph_theory)#:~:text=A%20different%20graph%20invariant%2C%20the,drawn%20simultaneously%20with%20straight%20edges.