Medium preparation, using calcium hydroxide instead of potassium hydroxide
As @Chris said, $\ce{K+}$ salts are soluble in water whereas plenty of $\ce{Ca^{+2}}$ salts such as $\ce{CaSO4}$ and $\ce{Ca3(PO4)2}$ are insoluble in the aqueous medium. Therefore it leads to the precipitation of phosphate-containing proteins and sulfate-containing metabolites. In addition, $\ce{Ca^{+2}}$ forms coordination complexes with calcium-binding proteins which participate in discrete signaling pathways. Interfering with this system could lead to subtle and irreproducible results. $\ce{K+}$ ion mostly participates in creating a voltage gradient across the membrane, nothing more.
Moreover, ionic strengths should be considered. $\ce{Ca^{+2}}$ is a divalent cation. A solution of $\ce{KOH}$ and a solution of $\ce{Ca(OH)2}$ with the same pH would definitely have different ionic strengths. Therefore despite achieving the same pH of the medium, ionic strengths vary depending on the salt used. Abnormal ionic strengths affect the structure of proteins in many ways. In general, it has been proven that changes in $\ce{K+}$ concentration lead to less severe outcomes compared to $\ce{Ca^{+2}}$.
EDIT: $\ce{Ca(OH)2}$ forms precipitates in an aqueous solution because it is barely soluble in water ($K_{sp}$ $\pu{=5.5 \times 10^{-6}}$). The pH of a saturated solution of $\ce{Ca(OH)2}$ is 12.347. To achieve other pHs one must add strong acid or base (a standard $\ce{KOH}$ solution perhaps) or use dilution; so as you can see, introducing $\ce{Ca(OH)2}$ just leads to unnecessary complexity, best to be avoided. Also, $\ce{Ca(OH)2}$ leads to the formation of $\ce{CaCO3}$ in the presence of atmospheric $\ce{CO2}$, making it even a less appropriate material for the preparation of the medium. Calculations using $K_{sp}$ shows a saturated $\ce{Ca(OH)2}$ solution has pH = 12.347 and [$\ce{Ca^{+2}}$] =  0.01111