This is basically a constraint satisfaction problem, I would not expect simulated annealing to get anywhere close to optimal in a reasonable amount of time. However, it might get you to a suboptimal solution quite quickly, so you could potentially terminate whenever you are out of time.
That said, if you want to solve this optimally the best way by far would be to use a CSP solver of some sort. I coded this up in IBM CPLEX OPL, which get's compiled into an integer linear program (ILP) and solved by the CPLEX solver. If you are in academia you can get a free copy of CPLEX and if you are not you can do a very similar thing in GLPK. You can also terminate many ILP solvers after a fixed amount of time and get the best solution so far.
Also, there are a number of speed ups you can do for this particular set up. First of all the green nodes can just be removed from the problem, they are always lined up along the top and left edges and if you will always have such a large number of green compared to red/blue it would never make sense to put a blue or red on those edges. These changes allowed the solver, for a 1000x1000 grid, to get within 7% of optimal in under 10 seconds, but is still hasn't found the actual optimal assignment after 15 minutes.
Here is the OPL code for a 100x100 grid with 100 green, 50 red, 50 blue, if you are interested.
using CPLEX;
dvar int grid[0..102][0..102][0..2] in 0..1;
minimize (sum(i in 1..101, j in 1..101, k in 0..2) grid[i][j][k]*(i*i + j*j));
subject to {
// edge conditions so I can always index i-1 and i+1 in all cases
forall(i in 0..102) (sum(j in 0..2) (grid[i][0][j] + grid[i][102][j])) == 0;
forall(i in 0..102) (sum(j in 0..2) (grid[0][i][j] + grid[102][i][j])) == 0;
// only one color per cell
forall(i in 1..101, j in 1..101)
(sum(k in 0..2) grid[i][j][k]) <= 1;
// 50 red
sum(i in 1..101, j in 1..101) grid[i][j][0] == 50;
// 100 green
sum(i in 1..101, j in 1..101) grid[i][j][1] == 100;
// 50 blue
sum(i in 1..101, j in 1..101) grid[i][j][2] == 50;
// green must be on the edge (not on not-edge)
forall(i in 2..100, j in 2..100)
grid[i][j][1] == 0;
// red must be next to another red
forall(i in 1..101, j in 1..101)
(1 - grid[i][j][0]) + grid[i+1][j][0] + grid[i-1][j][0] + grid[i][j+1][0] + grid[i][j-1][0] >= 1;
// blue cannot be next to another blue
forall(i in 1..101, j in 1..101)
(1-grid[i][j][2]) + (1-grid[i+1][j][2]) >= 1;
forall(i in 1..101, j in 1..101)
(1-grid[i][j][2]) + (1-grid[i-1][j][2]) >= 1;
forall(i in 1..101, j in 1..101)
(1-grid[i][j][2]) + (1-grid[i][j+1][2]) >= 1;
forall(i in 1..101, j in 1..101)
(1-grid[i][j][2]) + (1-grid[i][j-1][2]) >= 1;
}
Here is the optimal placement it solved in about 10 seconds on a 3.05Ghz machine
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