Let $T$ be a transcendence basis of $\Bbb R$ over $\Bbb Q$ and $F=\{A_n\subset T\colon n\in\Bbb N\}$ be a family of pairwise disjoint sets. Now, $$C=\{pA+q\colon A\in F \, and \,p,q\in Q\setminus\{0\}\}.$$
Q. Are all the sets in $C$ still pairwsie disjoint algebraically independent sets?
My answer is yes.
To see each set in $C$ is algebraically independent, let $pA+q\in C$, assume is not algebraically independent so there are algebraically dependent between their elements. This means there exist $\{pa_1+q,pa_2+q,\dots,pa_n+q\}\subset pA+q$ and a nonzero polynomial $f$ such that $f(pa_1+q,pa_2+q,\dots,pa_n+q)=0.$ Then we will have a nonzero polynomial, say $g$ such that $g(a_1,a_2,\dots,a_n)=0$ which is a contradiction since $\{a_1,a_2,\dots,a_n\}\subset A.$
To see all the sets in $C$ are pairwise disjoint. For contradiction, let $p_1 A+q_1, p_2 B+q_2\in C$ and $ (p_1 A+q_1)\cap (p_2 B+q_2)\neq\emptyset.$ Then, there exist at least $x\in (p_1 A+q_1)\cap (p_2 B+q_2)$ so $x=p_1a+q_1$ and $x=p_2b+q_2$. This gives us $$p_1a_1-p_2b_1+q_1-q_2=0$$ which is again algebraically dependent relation between $a_1$ and $b_1$ which is impossible since $\{a_1,b_1\}\subset A\cup B$ and $A\cup B$ is algebraically independent set.
EDIT. The case $A=B$ is Tomasz's answer below.
Is that right?
The solution is almost correct. The only issue I see (besides switching from $a,b$ to $a_1,b_1$ at the end) is that algebraic independence of $a$ and $b$ does not follow from $\{a,b\}\subseteq A\cup B$.
It follows from $A\cap B=\emptyset$ if $A\neq B$, and is simply false if $A=B$ (because then you might have $a=b$). But this case is also fairly easy to rule out for different reasons.