Numerically on the Koch curve, it seems to quickly converge. Symbolically, it is tedious to figure out any close formula. At stage $n$, rolling the unit sphere on the $(x,y)$-plane along $[0,\alpha]\times\{0\}\times\{0\}$ with $\alpha=3^{-n}$ rotates it around the $y$-axis with the matrix $A := A(\alpha):=\left(\begin{smallmatrix}\cos\alpha&0&\sin\alpha\\0&\;1\;&0\\-\sin\alpha&0&\cos\alpha\end{smallmatrix}\right)$. We only need to compose it with the $\beta:=\pi/3$ rotation around the $z$-axis whose matrix is $B:=\left(\begin{smallmatrix}\cos\beta&\sin\beta&0\\-\sin\beta&\cos\beta&0\\0&0&\;1\;\end{smallmatrix}\right)$ according to the recurrence
$$R_{n,0} := A(3^{-n})\;,\;\;R_{n,k+1} := R_{n,k}BR_{n,k}B^{-2}R_{n,k}BR_{n,k}\,.$$
The net rotation is $R_{n,n}$. In PARI/GP with 512-bit precision we instantaneously get $R_{100,100} $. Its first 20 decimals digits stabilize starting with $R_{56,56}$. Here they are :
$$\begin{pmatrix} \;\;\;0.54420\,52646\,18151\,51477 &  \;\;\;0.07439\,24346\,27331\,14793 & \;\;\;0.83564\,72914\,04756\,45330\\
-0.07439\,24346\,27331\,14793 & \;\;\;0.99641\,61277\,92991\,15095 & -0.04025\,74955\,03816\,29478\\
-0.83564\,72914\,04756\,45330 & -0.04025\,74955\,03816\,29478 & \;\;\;0.54778\,91368\,25160\,36382\end{pmatrix}$$
Does anybody recognize something there?