In a sense, there "aren't enough" unit quaternions to describe translations. Let me try to make this more precise, if not completely rigorous.
Conventionally, we can describe rotations using the set of unit quaternions
$H = \{a+bi+cj+dk: a,b,c,d\in\mathbb{R}, a^2+b^2+c^2+d^2=1\}$; each such quaternion can be written in the form $\cos(\frac\alpha 2) + \sin(\frac\alpha 2)(xi + yj + zk)$, representing a rotation by angle $\alpha$ around the axis given by $(x,y,z)$.
Just as the unit circle in $\mathbb{R}^2$ is described by the equation $a^2+b^2=1$, and the unit sphere in $\mathbb{R}^3$ is described by the equation $a^2+b^2+c^2=1$, the above description of $H$ identifies it as the unit hypersphere in $\mathbb{R}^4$. It is thus a compact space, which means that it is "small" in a certain precise sense. As a consequence of this, if you choose any sequence of unit quaternions, then it will have a convergent subsequence - that is, you can throw out enough points from that original sequence and thereby obtain a sequence which converges to a unit quaternion.
Now suppose we choose a way to describe rotations and/or translations by means of unit quaternions. This means that to each quaternion $q$ we associate a rotation or translation $f(q)$. In order to be useful for computation, this association should be a homomorphism: it should satisfy the rule $f(q_1q_2) = f(q_1)\circ f(q_2)$. This ensures that $f$ relates the multiplication of quaternions to the composition of transformations, which is what we usually mean by the quaternions "describing" a set of transformations.
This map $f$ should also be continuous: if $q_1,q_2,\ldots$ is a sequence of unit quaternions which converges to some unit quaternion $q$, then the sequence $f(q_1),f(q_2),\ldots$ should converge to $f(q)$. We impose this continuity condition to avoid strange behaviour, and because in practice just about any reasonable $f$ you can come up with will be continuous.
I claim that no matter how you choose such an $f$, it cannot produce any translation. On the contrary, suppose that we had some $q$ such that $f(q)$ is a translation. Precisely, suppose that $f(q)$ is the translation "add $v$" where $v$ is some nonzero vector in $\mathbb{R}^3$. Since $f$ is a homomorphism, it follows that $f(q^2)$ is the translation "add $v$, then add $v$ again", i.e. "add $2v$". Likewise, for any $n>0$ we have that $f(q^n)$ is the translation "add $nv$".
Consider the sequence $q,q^2,q^3,\ldots$ of unit quaternions. Since $H$ is compact, there are positive integers $n_1< n_2< n_3\cdots$ such that the sequence $q^{n_1}, q^{n_2}, q^{n_3}, \ldots$ converges to a unit quaternion $\bar q$.
Now since $f$ is continuous, it follows that the sequence "add $n_1v$", "add $n_2v$", "add $n_3v$", $\ldots$ converges to $f(\bar q)$, which is some transformation of $\mathbb{R}^3$. But this sequence can't possibly converge to anything, because the sequence of vectors $n_1v,n_2v,\ldots$ shoots off to infinity and so the corresponding sequence of translations does, too. By this contradiction, we conclude that $f(q)$ could not have been a translation.
Best Answer
To apply a subsequent rotation corresponding to $q_1$ and translation corresponding to $T_1$ to your $v'$, you'd just do it in terms of quaternion algebra: you'd get
$v''=q_1v'q_1^{-1}+T_1=q_1(qvq^{-1}+T)q_1^{-1}+T_1=(q_1q)v(q_1q)^{-1}+q_1Tq_1^{-1}+T_1$
Thus, as you say, it is possible to do the computations on the transformations cumulatively, and then apply it to the data, to avoid cumulative distortion in the data.
(edited to correct original typo in formula)