Solution:

Let the roots be $r,s.$ Then: $$\begin{align*}&x^2+kx+36 \\ &= (x-r)(x-s) \\&= x^2-(r+s)x+rs \\&=0 \end{align*}$$ And so, $rs = 36$ and $r + s = -k.$

Therefore, we need $r$ and $s$ distinct such that $rs = 36.$ All the possible factor pairs are $$\pm\{1,36\},\pm\{2,18\},\pm\{3,12\}$$ $$\text{and }\pm\{4,9\}.$$

Each of these unordered pairs produces a unique value for $k,$ so there are $8$ possible values for $k.$

Thus, B is the correct answer.