Here is a basic observation :
On page 68 of his article Reductive groups over local fields, Jacques Tits writes:
All types of groups listed (...) exist over an arbitrary [non-archimedean local] field (...). The central isogeny class corresponding to a given name is always unique except in the following cases. (...)
Let us ignore those exceptions, and assume that we are given a type, say $ t $, whose isogeny class over any non-archimedean local field is unique. Then the situation is reminiscent of the split situation, where for any given type, one proves the existence and uniqueness (up to isogeny) of this type of split groups over any field.
This motivates the following question:
Let $ t $ be a type of algebraic group as in the previous paragraph. Does there exist an algebraic group defined over $ \mathbb{Q} $ whose base change to $ k $ is of type $ t $, for all non-archimedean local field $ k $ of characteristic $0$ ?
(see grghxy comments why we restrict to char. $0$ and why we then can use $\mathbb{Q}$ instead of $\mathbb{Z}$).
EDIT : I edited the question to take into account all comments of grghxy.
Note that so far, the only given obstruction comes from a non-trivial Galois action on the Dynkin diagram. But looking at the classification, one can see that if the Galois action is non-trivial, then the type of algebraic group has more than one isogeny class (and the converse is true except for the form $SL_n (D)$).
So, the question remains : for types that have a unique isogeny class, is there any obstruction to being defined over $\mathbb{Q}$ ?
Here is the list of such non-split types. They are in fact uniquely determined by their Tits index $X_{n,r}$ (where $X_n$ is the absolute type, and $r$ the relative rank): $B_{n,n-1}$, $C_{2n-1,n-1}$, $C_{2n,n}$, $D_{2n+1,n-1}$, $E_{6,2}$, $E_{7,4}$
