Here is a heuristic argument suggesting that additive squares are unavoidable.
Consider the alphabet {1,2,..,k}.
We forbid additive squares uu such that 1 <= |u| <= n.
A heuristic upper bound on the growth rate is k*(1-1/k)*(1-1/(2k))*..*(1-1/(n*k)).
By appending a letter to the right, we have k possibilities, and the word ends with uv such that |u| = |v| = i and 1 <= sum(u) = sum(v) <= i*k with "probability" 1/(i*k).
We also assume "independence".

This expression is smaller than 1 for n > k^k and goes to 0 when n goes to infinity.
By taking the logarithms, we see that this is because Sum(1/i) diverges.
The corresponding expressions for additive cubes and abelian squares are Sum(1/i^2) and Sum(1/i^(k-1)), which converge.