More precisely:
A multivariate function $f(x)$, $\mathbb{R}^n \mapsto \mathbb{R}$, is monotone increasing if $\forall x,y \in \textbf{dom}$ $f$ if $\forall i, x_i \leq y_i \implies f(x) \leq f(y)$ and similarly for monotone decreasing.
Is such a function always quasiconvex or quasiconcave?
This is false, I came up with the following counterexample:
$f: R^2 \to R$ s.t $f(2,0)=1$, $f(0,2)=1$ and $f(1,1)=2$. For all other points in the $(x,y)$ plane, define $f(x,y)$ to be $0$ if $y < 2-x$ and $f(x,y)=2$ if $y \geq 2-x$ (obviously with the exception of the 3 points i singled out).
Note first that $f$ defined in this way is not quasi-convex, since $f(\frac{1}{2}(2,0)+\frac{1}{2}(0,2))=f(1,1)=2 > \max\{f(2,0),f(0,2)\}=1$ On the other hand $f$ is "monotonic" as per your definition, it's easier to check this graphically I think.
It helped me to draw a point $(c,d)$ in the $x,y$ plane and consider the set of points $(a,b)$ such that $f(a,b) \leq f(x,y)$ because of monotonicity of $f$ these were exactly the points "below" AND to the "left" of $(c,d)$, and this observation helped me make the counterexample.