This is too long for a comment. If you want a single $n=n(k,\varepsilon)$ and do not care about the dependency on the parameters ($\varepsilon$ is the largest diameter of the pieces), then here is one way to do it. This is only a half-answer, since it does not provide a well-known reference.
The result is obvious for $k=0$, so we assume $k\geq1$. Choose some $r\geq100k$.
- Cut the sphere along half-hyperplanes
$$\{-\sin(\theta_i)x_1+\cos(\theta_i)x_2=0,\cos(\theta_i)x_1+\sin(\theta_i)x_2\geq0\}$$
for $\theta_i$ very close to $i\pi/r$ (say $\pi/4r$-close), $1\leq i\leq2r$, in such a way that the volumes of the pieces $\{C_i\}$ you created are rational fractions of the total volume of the sphere. These are convex pieces, with respect to the spherical geodesics. (This is a bit ridiculous since we can just choose $\theta_i=i\pi/r$, but bear with me.)
- Cut every $C_i$ along half-hyperplanes
$$\{-\sin(\theta_{ij})x_1+\cos(\theta_{ij})x_3=0,\cos(\theta_{ij})x_1+\sin(\theta_{ij})x_3\geq0\}$$
for $\theta_{ij}$ very close to $j\pi/r$, so that all the pieces $\{C_{ij}\}$ thus created still have rational volume. This is possible since the projection of the interior of $C_i$ on these components is connected ($C_i$ is convex). Iterate for all pairs of dimensions.
- Find a common denominator $q$ for all these ratios, i.e. every piece $C_{i_{11},\ldots,i_{k+2,k+2}}$ has volume $p_{i_{11},\ldots,i_{k+2,k+2}}\operatorname{vol}(S^{k+1})/q$. For each of them, consider an isometry $\phi:R^{k+2}\to R^{k+2}$ such that $\phi^{-1}(0)$ belongs to its interior, and cut the piece along half-hyperplanes of the form
$$\{-\sin(\theta)\phi(x)_1+\cos(\theta)\phi(x)_2=0,\cos(\theta)\phi(x)_1+\sin(\theta)\phi(x)_2\geq0\}$$
so that every piece has volume $\operatorname{vol}(S^{k+1})/q$.
These $q$ pieces all have the same volume. We can bound their diameter as follows. Let $x,y$ be two elements of the same piece, and suppose they are at (spherical) distance at most $\pi/2$. Since they are unit, the norm $|x\wedge y|$ of their exterior product is the sine of the angles between the two, i.e. $|x\wedge y|=\sin(d(x,y))$ for $d$ the spherical distance. Moreover,
$$ |x\wedge y|^2 = \sum_{i<j}(x_iy_j-x_jy_i)^2 < \binom{k+2}2\sin\left(\frac{3\pi}{2r}\right)^2. $$
We find a crude estimate $d(x,y)\leq4(k+3)/r$. This is smaller than $\pi/2$, so by convexity the diameter has to be at most $4(k+3)/r$, which can be made arbitrarily small by choosing $r$ very large.