NGP in Finkelstein reaction
You are thinking too far ahead for a simple reaction. First of all, $\ce{I-}$ is not a very strong base to abstract a protone from alcohol. $\ce{I-}$ is the conjugate base of very strong acid, $\ce{HI}$. Yet, $\ce{I-}$ is a very good nucleophile. Therefore, the condition given ($\ce{NaI}$ in anhydrous acetone; $\ce{NaI}$ dissolves in anhydrous acetone and therefore exists as $\ce{I-}$ in solution) is suitable only for nucleophilic substitution reaction and also favored $\mathrm{S_N2}$ mechanism. Since the byproduct form ($\ce{NaCl}$) is not soluble in acetone, the reaction favors only the forward reaction:
$$\ce{R-Cl + Na+ + I- -&gt; R-I + NaCl_{(s)}}$$
Since it is a $\mathrm{S_N2}$ reaction, inversion is the result ($(1R,2S)$ to $(1R,2R)$ with closed to $100\% e.e.$).
However, note that there is a neighboring group participation (NGP) possible in 1,2-halohydrins in certain reactions (Ref.1). As OP suggested, if NGP is contributed here (without strong base I doubt it though), the product from NGP is going to be the retention product with $(1R,2S)$ configuration. Thus, total $\%e.e.$ of final product is $\lt 100\% e.e.$.
References:

T. A. Geissman, Richard I. Akawie, &quot;Rearrangements of Halomagnesium Derivatives of Halohydrins,&quot; J. Am. Chem. Soc. 1951, 73(5), 1993–1998 (https://doi.org/10.1021/ja01149a029).