The question should be reasonably selfcontained: do MHM's form a stack? This is wellknown for perverse sheaves, proved already in [BBD(G)]; does it hold for mixed Hodge modules? In other words, if I have a variety $X$ over the complex numbers, an open cover $(U_i)$ and mixed Hodge modules on all the $U_i$ with gluing isomorphisms on overlaps with the obvious cocycle condition, then do I get a unique MHM on $X$? I would like this in the analytic category but a reference for algebraic varities over the complex numbers is likely to satisfy me.
This is an extremely partial (and sketchy) answer that I haven't fully thought through. But, I am essentially just running the proof for perverse sheaves. Let me first construct the required MHM if the cover just consists of two elements. In this case the required MHM can be define as the cone in the MayerVietoris distinguished triangle. Now use induction to prove the statement for a countable cover. For arbitrary covers I am not quite sure how to proceed. On the other hand, all I am saying is that define the MHM as the cohomology of the Cech complex given by your data. Unless I am missing something, this (modulo some technicalities with finiteness assumptions) gives the construction. Now for uniqueness, probably one can just observe that the usual isomorphism that shows uniqueness for perverse sheaves is a morphism of MHM. Since it is an isomorphism on underlying perverse sheaves, it is an isomorphism on the MHMs. In the case of a 2element cover this also follows from the fact that there are no negative Exts between MHMs (which leads to the cone in the MayerVietoris triangle being unique). 

