Consider the inclusion $k\subset A$ of the field $k$ in the domain $A$ and the fraction field $K=Frac(A)$ of $A$.
Obviously if a family $(a_i)_{i\in I}$ of elements $a_i \in A$ is algebraically independant over $k$ it will remain algebraically independant in $K$.
Consider however a family $(\alpha _i) _{i \in I}$ of elements $\alpha _i \in K$ algebraically independent over $k$.
To my puzzlement, I can't construct from it an algebraically independent family $(a_i)_{i\in I}$ of elements $a_i \in A$. Although my real question is whether it is possible to actually construct such a family in a natural way, I'll ask something more precise:

Precise question Given the $k$- algebraically independent set $(\alpha _i) _{i \in I}$ in $K$, does there exist in $A$ some $k$- algebraically independent set $(a_i)_{i\in I}$ ( with the same index set $I$) ?

The answer is "yes" if $A$ is finitely generated over $k$., thanks to E.Noether's normalization theorem. Interestingly the proof of that theorem is not purely field-theoretic, since it makes use of Krull dimension.

NB I'm not sure (despite the title of the question!) that I know what the transcendence degree of $A$ is: the "correct" definition might follow from the answers to this question!

Edit a-fortiori has proved in his comment that the anwer to the "Precise question" is yes, and that as a consequence the only reasonable definition of transcendence degree of $A$ is that it equals the transcendence degree of $K$.
I now think that it is impossible to naturally associate to the $k$- algebraically independent family $(\alpha _i) _{i \in I}$ in $K$ a $k$- algebraically independent family $(a_i)_{i\in I}$ in $A$, even though we now know thanks to a-fortiori that such a family exists. For example, if $X, Y$ are algebraically independent over $k$, and we take the family of just one element $\alpha=\frac {X}{Y}\in k(X,Y)$, which transcendental element in $k[X,Y]$ should we choose?! It would be great if someone could come up with a rigorous statement of the impossibility of a natural choice.