Does gradient descent work for tabular Q learning?

Yes, it is possible; you are close, but not quite there.

You lost a gradient in your equation; it should be: $$Q_{k+1}(s,a) = Q_k(s,a) - eta left(Q(s,a)-(r+gammamax_{a'}{Q(s',a')})right)left(left.frac{d~Q}{d~theta}right|_{(s,a)} - gamma left.frac{d~max_{a'}Q}{d~theta}right|_{(s')} right)$$

Which does simplify a bit in the case of a tabular representation:

$$Q_{k+1}(s,a) = Q_k(s,a) - eta left(Q(s,a)-(r+gammamax_{a'}{Q(s',a')})right)left(1 - gamma left.frac{d~max_{a'}Q}{d~theta}right|_{(s')} right)$$

Problems may arise if $s=s'$ and $a=a'$, because your update will be $0$ (which it shouldn't). It's also not a good idea to try and differentiate the $max$ function.

You can do the "double deep q-learning trick" and introduce $theta_textrm{old}$ to estimate $Q(s',a')$, i.e., use the q-table from the previous step. This will make the other gradient dissapear, and you are indeed left with q-learning:

$$Q_{k+1}(s,a) = Q_k(s,a) - eta left(Q(s,a, theta)-(r+gammamax_{a'}{Q(s',a', theta_textrm{old})})right)$$

In this case, the loss will be

$$hat L(theta)= frac12 left(Q(s,a,theta)−(r+gamma max_{a′}Q(s′,a′, theta_textrm{old}))right)^2$$