Here is one way to get a theorem about this situation:

Although $M'$ may not be invertible, you can always adjust by a scalar $\lambda \in K^\times$ to ensure that $(\lambda M)'$ is invertible. This can be seen as follows: let $\{\alpha_1, \dots, \alpha_n\}$ be a basis for $K/\mathbb{Q}$. Write $M = \alpha_1 M_1 + \dots + \alpha_n M_n$ for rational matrices $M_i$. Now $\mathrm{det}(x_1 M_1 + \dots + x_n M_n)$ is a polynomial in $\mathbb{Q}[x_1,\dots,x_n]$ which has non-zero value at the point $(\alpha_1, \dots, \alpha_n) \in K^n$, and so is non-zero. Since $\mathbb{Q}$ is infinite, there is also a rational point $(\beta_1, \dots, \beta_n) \in \mathbb{Q}^n$ at which it is non-zero. Now by the nondegeneracy of the trace form, we can pick a $\lambda \in K$ such that $\mathrm{tr}_{K/\mathbb{Q}}(\lambda \alpha_i) = \beta_i$ for all $i$. Then $\mathrm{det}(\mathrm{tr}_{K/\mathbb{Q}}(\lambda M)) \neq 0$ as desired.