Let $\{ z_1,\ldots,z_n\}$ be any finite set, and $$L_k(z)=\prod_{j\neq k}(z-z_j).$$ Then $L_k,\; 1\leq k\leq n$ are linearly independent since the set of their linear combinations consists of all polynomials of degree $\leq n-1$ (Lagrange interpolation formula says that), and the space of all polynomials of degree $\leq n-1$ is of dimension $n$.
Now we can easily prove (by contradiction) that for every $\{ z_1,\ldots,z_n\}$ there exists $\epsilon>0$ such that the disks $\{ z:|z-z_k|<\epsilon\}$ have your desired property.
I understand that this is not what you desire; you want a conformally invariant "separation parameter".
However this argument shows that if the "separation parameter" of a finite collection of disks has the property that when it tends to infinity, the disks shrink to points, then your statement about linear independence is true.
The separation parameter that you propose does not have this property, and in fact it has another important drawback: it can stay $>$ some positive number while some of your disks are arbitrarily close to the unit circle (which is the boundary of one of the disks of your family). And your separation parameter depends on which disk you choose to send to the "upper hemisphere". So your "separation parameter can be very large, while some disks are very close to each other.
I can propose a separation parameter which will have the desired property: it is the minimum (over $1\leq k\leq n$) of the reciprocal extremal length of the family of curves which separate one disk $D_k$ from the rest.
When this tends to $\infty$, this means that all these extremal lengths are small, so the disks are well separated, and must shrink top points.
For the definition and properties of extremal length, see any of the two books my Ahlfors, Conformal invariants, or Lectures on quasiconformal mappings.