Why is cut-off regularization is not Lorentz invariant? Why is it said that the cut-off regularization is not a Lorentz invariant regularization method?
 A: The UV cutoff procedure that is used to regulate Feynman diagrams must be chosen so that the region of integration involving large values of $p^\mu$ (all components individually) is cut-off. A Lorentz invariant cut-off such as $p^2 < \Lambda^2$ does not achieve this. 
However, what we can do is to Wick rotate $p^0 \to - i p^0$ and then impose a cut-off on $p_E^2 < \Lambda^2$. The latter cut-off does exactly what we require it to do. In fact, this Wick rotation is convenient for other purposes. In any case, it is clear that such a cut-off is not Lorentz invariant. 
A: Reason 1: Rotational Invariance Is Broken
Consider the following integral: $$\int d^4l\, F(l^2)\,l_\mu $$ in general, that is don't yet think about "cut-off" or anything. This integral is zero because $l_\mu$ is not rotationally invariant while everything else ($d^4l\, F(l^2)$) is. Now, when one imposes a "cut-off" on $l$ which is equivalent to moving from a continuous space-time to a discrete lattice. In the continuous space-time suppose $l$ is directed along a certain direction, you rotate the whole continuous space-time such a way that now in this new rotated frame the original vector $l$ becomes $-l$ and as $d^4l F(l^2)$ is invariant the sum of $l+(-l)$ gives zero. What happens in the discrete case is that you are not allowed to rotate by any angle you wish, only a finite discrete set of angles are allowed under which your $d^4l F(l^2)$ remains invariant, and so using this restricted set of angles you may not find a frame where the vector becomes $-l$ so that it cancels $l$ !
Reason 2: Translation Symmetry Is Broken Too
A translation $l\mapsto l+q$ for any $q$ is not possible, because you have fixed lattice axes and you are allowed to move only along them and by an integral multiple steps of the lattice spacing.
So the summary is the following:
A momentum cut-off $\Lambda$ let's say, is equivalent to a lattice which does not enjoy the Poincar`e symmetry.
