Although the decomposition is not unique, the diagonal form of $A$ (that is $\Lambda$) is unique up to the order of the diagonal entries. Thus, if $ A = Q\Psi Q^T$ is another decomposition of $A$ with $\Psi$ diagonal and $Q$ orthogonal, then $\Psi$ and $\Lambda$ are related by permuting rows and columns; that is, there is a matrix $R$ which is obtained by permuting the columns of the identity matrix (and in particular, $R$ is orthogonal) such that $R\Psi R^{-1} = R\Psi R^T = \Lambda$.
Note that $A = U\Lambda U^T = U(R\Psi R^T)U^T = (UR)\Psi(UR)^T = Q\Psi Q^T$.
It is then straightforward to check that $R\Psi^{1/k}R^{T} = \Lambda^{1/k}$, and that $(UR)\Psi^{1/k}(UR)^T = Q\Psi^{1/k}Q^T$.
So if you pick the decomposition $Q\Psi Q^T$ instead of $U\Lambda U^T$, then the positive definite $k$th root of $A$ you get will be $B = Q\Psi^{1/k}Q^{-1}.$
But $U\Lambda^{1/k}U^T = U(R\Psi^{1/k}R^T)U^T = (UR)\Psi^{1/k}(UR)^T = Q\Psi^{1/k}Q^T;$ that is, the matrix you get is actually equal to the one you got originally.