Can truth tables be used in non-boolean algebra to derive functions?
There are two possible ways to define a non-boolean algebra.


Stick with $\mathbb{F}_2 = \{0,1\}$ but choose base operations other than $\{\land, \lor, \lnot\}$.
Change the base set entirely.


For the whole concept of simplification to word similarly to boolean algebra, we need two binary operations with associativity, commutativity and distributivity. Fields are the natural choice and, I think, the only one. Commutative rings might work but I suspect you get into trouble composing functions in a regular way if you don't have multiplicative inverse elements.

Infinite fields do not make sense here -- you can not draw an infinite table. So finite fields remain, of which there is exactly one per size (up to isomorphy).

So let us, for example, consider $\mathbb{Z}/5$ (the smallest field over $\mathbb{Z}$ with more than two elements and interesting inverses). Now try and formulate a function in a way that looks like a boolean function provided by Karnaugh maps, that is in additive (disjunctive) or multiplicative (conjunctive) normal form. We get to the core question:


  What is the complement of a given element in a non-binary algebra?


The simple logic "if it's not one it's the other" does not work here. Note that the boolean complement does not correspond to the additive nor the multiplicative inverse, so there no hope there. So let's try and define a "complement"; say $\overline{a} = a + 1$ (which contains the boolean complement as special case). We note with unease that this new complement is not symmetric. I expect symmetry to be of essence; reading off from the Karnaugh map seems to rely on it.

So there is no way to get there with an odd number of elements; let's pick one with an even number (so we can pair elements for a complement relation). Now things become messy and I'll stop. After all, Wikipedia says that (emphasis mine)


  the solution can be found by eliminating extra variables within groups using the axioms of boolean algebra


which characterise boolean algebra. Therefore -- provided that all axioms are needed for Karnaugh maps to work -- the concept does not work for non-boolean algebras.