How can the class of tail recursive functions be compared to the classes of PR and R?
Every computable function can be expressed in continuation-passing-style, in which all calls are tail-calls.

The trick is to add a "continuation" parameter to every function. Instead of making a non-tail-call to a function, you make a tail call to that function with a modified continuation, describing what to do with the result. All instances where a value is directly returned (such as recursion base cases) are replaced by calling the continuation in tail-position with the result as an argument.

Transformation from the lambda calculus into CPS can be done mechanically. So, $TR=R$.



EDIT: addressing the comments about higher-order functions:

Tail-calls are almost always discussed in the context of higher order functions and the lambda calculus. So the problem is, what precisely is our definition of $TR$?

You can certainly add a wrapper around a CPS function to give it the type $\mathbb{N}^n \to \mathbb{N}$, by giving it an initial continuation of $\lambda k \ldotp k$. If higher-order functions are allowed internally, then the result that $TR=R$ still holds.

If higher order functions aren't allowed internally, what is our definition of $TR$? If we define it in the same way as $PR$, then it is going to only contain primitive recursive problems by definition (since it's just the restruction of $PR$ to tail-recursion). If we add $\mu$ for infinite search, I think we're just going to get $R$, since we can encode higher-order functions using integers. So, I'm not sure there's a meaningful question to be asked in the non-higher-order case.



EDIT 2: 

As for the class of first-order functions that only allow tail recursion, with Constant, Successor, Projection and Composition functions, and extension by tail recursion:

h(x1 ... xn) = 
  if c(x1 ... xn) = 0 then
    h(g1(x1), ..., gn(xn))
  else
    f(x1, ..., xn)


where $c$, $g_i$ and $f$ are all tail-recursive functions, I think we can prove that it's Turing Complete, by solving Post's Correspondence Problem, which is undecidable but semi-decidable:

Assume that we've got nice functions for dealing with strings encoded as integers, with concatenation, etc.

Let $pcpInst(k, n)$ be a function which takes an integer $k$ and returns the $k$th string over the alphabet $\{1, \ldots, n \}$.

Let $c(k, x_1, \ldots, x_n) = $ be a function, where $k$ is an integer, and each $x_i$ is a pair containing two strings over a binary alphabet. Thus function does the following:


Computes $k_1 \cdots k_p = pcpInst(k,n)$, the $k$th possible PCP solution indices.
Constructs $s_1=\pi_1(x_{k_1}) \cdots \pi_1(x_{k_p}))$. This is the string we get by concatenating the first string of the arguments indexed by our $k_i$ sequence. We define $s_2$ with $pi_2$ similarly.
Return $0$ if $s_1 \neq s_2$, return $1$ otherwise.


Now, we'll define our function to solve the a PCP instance with $n$ strings:


$h(k, x_1, \ldots, x_n) = h(S(k), x_1, \ldots, x_n)$ if $c(k, x_1, \ldots, x_n) = 0 $
$h(k, x_1, \ldots, x_n) = 0$ otherwise


Now we define $h'(x_1, \ldots, x_n) = h(0, x_1, \ldots, x_n)$.

It is clear to see that $h'(x_1, \ldots, x_n)$ returns 0 if and only if there is a solution to the correspondence problem defined by pairs of strings $x_1, \ldots, x_n$. If there is a solution, we eventually iterate to it by increasing $k$, and return $0$ when our $c$ function returns 1. If there is no solution, we never return.

The trick here is ensuring that $c$ is itself tail-recursive. I am fairly confident that it is, but that proving so would be tedious. But since it is performing simple string manipulations and equality checks, I would be very surprised if it is not tail recursive.