Is $[\bar{\mathbb Q}:\bar{\mathbb Q}\cap\mathbb R]=2$? 
Is $[\bar{\mathbb Q}:\bar{\mathbb Q}\cap\mathbb R]=2$ ?

I think it is true that $\bar{\mathbb Q}\cap\mathbb C=\bar{\mathbb Q}$, because I've heard that the closure of the reals is $\mathbb C$. And $\bar{\mathbb Q}$ is a subfield of $\mathbb C$, so using this fact, we have $\bar{\mathbb Q}\cap\mathbb R$, intersection of $2$ fields, again a field.
but does the index necessarily have to be an integer as in the group theory ?
 A: By definition the index is the dimension of $\bar{\mathbb Q}$ as a vector space over $\bar{\mathbb Q}\cap \mathbb R$, so it is always an integer (if finite).
If we let $F=\bar{\mathbb Q}\cap \mathbb R$, it is not difficult to prove that every element of $\bar{\mathbb Q}$ can be written $a+bi$ with $a,b\in F$, so the dimension (and therefore the index) is indeed 2.
A: The real algebraic numbers form a field, yes. More generally, the intersection of two subfields of a field is again a field. 
A number is algebraic if and only if its real and imaginary part are algebraic. 
Thus one can write every algebraic number $a$ as $r_1 + i r_2$ with $r_1,r_2$ real algebraic numbers. 
The degree is thus $2$ as you suspect. 
The degree of the field extension is the dimension of the large field as a vector space over the smaller. It is thus a positive integer or an infinite cardinal, like the index of a subgroup. 
A: Yes, $[\overline{\mathbb Q}:\overline{\mathbb Q}\cap \mathbb R]=2$, since you attain the former from the latter by adjoining $i$.
The index is always either a positive integer or infinite, since it represents the cardinality of a basis of a vector space. 
