My first understanding was yes, but in fact, it depends.
First understanding: You can view them as IID, unless you have some knowledge about the secret $s$ and enough other variables. Take for example a Shamir Secret, a simpler secret sharing scheme which is a special case of this. A Shamir secret allows to reconstruct $s$ with any set of $k$ parts from $n$ shares. It is built by choosing a random $k-1$ degree polynomial such that $f(0)=s$. Any set of $k$ or more distinct points allows you to reconstruct this polynomial curve and thus learn it's value for $x=0$. Any set of $k-1$ points (except $(f(0),0)$) would be unauthorized and insufficient to reconstruct the secret. Since each coefficient of the polynomial (except $s$, that you have no knowledge of, including it's distribution) is chosen randomly, every set of point. Conditioning on $k-1$ shares doesn't change in any way the distribution of the $k^{th}$ share: it is still random.
But in fact it's not necessarily the case: only parts of these shares must be randomly distributed. Take these Shamir secrets: every share is a tuple $(x,f(x))$, and given their construction and a tuple $(x_1,f(x_1))$, you would expect the next tuple not to have $x_2 \neq x_1 $, so that even if $f(x_2)$ would be independent of $(x_1,f(x_1))$, $(x_2,f(x_2))$ would not. Conversely, you could choose $x$ as random every time, risking a possible $x_2 = x_1$, but making all the shares IID.
