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I am sorry, I don't know the answer to your question but I just realized that you can prove it using Gröbner basis. Let $E$ and $F$ be subfields of $K$ such that $I$ is generated by polynomials with coefficients in $E$ and in $F$, respectively. Then choose reduced Gröbner bases $G$ and $H$ of $I$ with respect to the same term ordering having all coefficients in $E$ and in $F$, respectively. Now both $G$ and $H$ are reduced Gröbner bases of $I$ also over $K$. Because of the unicity of the reduced Gröbner basis, we have $G=H$. Hence $I$ is generated by polynomials with coefficients in $E\cap F$.
I just remarked that I accidentally added the constraint "set(m>=0)". However, without this constraint YALMIP does not find (at least at this relatively low relaxation degree of 4) an optimal solution. And moreover, if you add the constraint "set(m<=0)", then YALMIP gives a lower bound of 11.7390 (at the same relaxation degree) for the true optimal value of the problem with the additional constraint m<=0. Henceforth, the solution I gave above should nevertheless be correct (numerically).
If the answer to your question were no, then by a result of Bre\v sar and Klep (Corollary 5.8, Tracial Nullstellens ̈\"atze, Notions of Positivity and the Geometry of Polynomials, Trends in Mathematics, 79–101) the difference of the two words would be a binomial in two non-commuting variables over the rational numbers which can be written as the sum of a non-trivial polynomial identity in two variables for $3\times 3$-matrices and a sum of commutators. In particular, the answer to your questions is yes if both words have length at most five.
