The set of tilings (which I'll call $T$) corresponds in number to the points on a line, so this essentially claims that $\lvert T\rvert=\lvert\mathbb R\rvert=\mathfrak c$. Instead of actually establishing a bijection, you can also establish two injections.
Upper bound on the cardinality
$\lvert T\rvert\le\lvert\mathbb R\rvert$ is easy: interpret $L,R,l,r$ as digits $0,1,2,3$ and the whole infinite sequence as a decimal fraction. Since the digit $9$ does not occur, you don't have to worry about things like $0.99\!\ldots=1$ so you know that every infinite sequence corresponds exactly to one real number. So you have an injective map from $T$ to $[0,1)\subset\mathbb R$.
Uncountability of sequences
$\lvert T\rvert\ge\lvert\mathbb R\rvert$ is harder. You could interpret rational numbers as fractions with base $4$, so every digit would correspond to one of the letters. You could also do the following preprocessing:
$$f(x):=\begin{cases}
L,g(x) &\text{if } x\in[0,1) \\
R,g(1/x) &\text{if } x\in[1,\infty) \\
l,g(-x) &\text{if } x\in(-1,0) \\
r,g(-1/x)&\text{if } x\in(\infty,-1]
\end{cases}$$
This will map any $x$ to the range $[0,1]$ which can then be used as input to the second function $g$ that represents that number in base $4$ and thus turns it into a sequence. You could write $1=0.333\!\ldots_4$ or tweak the definition of $f$ a bit (e.g. using $1-g(1/x)$ in the second case) to ensure that you only translate numbers from the range $[0,1)$ so that you can concentrate on the digits after the point.
Uncountability of point labels
So at this point, you have an injective function from $\mathbb R$ to infinite sequences in these four letters. But now you have a technical problem:
Not all possible sequences of the four symbols can be produced this way
You have to work out the exact conditions of which sequences can and which can not be produced in this way. My guess is that you can show that there are only countably many forbidden sequences, so the remaining sequences would still be uncountably many.
Uncountability of tilings
As Daniel reminded me in a comment, what we have worked to so far (at least if you manage to show the countability of the sequences which are not tilings) demonstrates that there are uncountably many sequences describing a pattern seen from a given tile. But there are infinitely many tiles from which you could label the pattern and it would still be the same pattern that should be counted only once.
Two sequences denote the same pattern if they agree after some point $k$. So you can generate alternate labels for the same pattern by subsequently modifying elements of the sequence, starting at the first. You have four options for the first sequence item, and for each of them four for the second, and so on. You could explore all possible sequence modifications in a breadth first search, therefore there are only countably many (i.e. $\lvert\mathbb N\rvert=\aleph_0$) of these.
So why does that not break the uncountability of $T$? I'm not sure how you feel, but (at least at the moment) a counter-argument works best for me. Suppose there were only countably many tilings in $T$, represented by any suitable sequence for each. Then you could use the common $\mathbb N^2$ iteration: Take the original label of the first tiling, then the original label of the second tiling followed by the first modification of the first tiling, then the original label of the thirs tiling followed by the first modification of the second tiling followed by the second modification of the first tiling, and so on. In the end you'd enumerate all point labels, so there could only be $\lvert\mathbb N^2\rvert=\lvert\mathbb N\rvert=\aleph_0$ of these. This is a contradiction to the uncountability of label sequences we showed before.
Now that I think about it, this only shows $\lvert T\rvert>\aleph_0$ but one would need the continuum hypothesis to deduce $\lvert T\rvert=\mathfrak c$ from this. So if anyone has a better explanation (which still works on an intuitive level) of why $\mathfrak c/\aleph_0=\mathfrak c$ feel free to edit this post, comment, or provide your own answer.
Here is an proof that for odd $n$, $n$ colors is not enough, from which it follows that we need $n+ 1$ colors.
Lets assume we need only $n$ colors. Place the W-polyomino anywhere, and color the cells that it covers from $0$ to $n - 1$.
Now place the W-polyomino so that it covers all the same cells except the one with color 0. The uncolored cell must be of color 0 (since we have only $n$ colors and all the other colors are used already).
Repeat with color 1.
We can continue in this way, but for this proof we need to go only this far.
Now put the W-polyomino so that it covers the first few cells with even color, but none with odd colors. The cells that are not colored yet must be colored with the odd colors, one of each (again, because we only have $n$ colors). In particular, one of the cells must be color 1. It cannot be the cell next to color 0, since it lies in the same W shape as the cell below which also has color 1. Therefor, one of the remaining cells must be color one. Those cells are marked orange in the figure below.
Finally, move the W-polyomino one cell down and one to the right. It now covers all three orange cells (one of which must be color 1), and another cell with color 1. This violates the condition of this being a legal coloring, which means, we cannot get away with only $n$ colors.
It is worth seeing how this proof fails for when the number of cells is even, and for other polyominoes resembling W-polyominoes, such as these:
There are two possible extensions that may be worth looking at.
The first is polyominoes similar to W-polyominoes (scaled copies of W-polyominoes).
My suspicion is that for $W_n$ scaled by $r$, we need $r^2(n+1)$ colors if $n$ is odd. The proof above does not work directly, but if we keep careful track of the extra colors, and analyse the cases, it works. What is interesting about this specific example is that it is usually easy to prove that if we need $C(P)$ colors for $P$, we need $r^2C(P)$ colors for a copy of P scaled by $r$ (provided it holds for rectangles, which is the other case where it is not always easy to prove).
The second is this type of polyomino:
L-polyominoes are interesting because they require so much colors compared to their area (the same as their bounding box); W-polyominoes are interesting because they can be quite large without having large Ls as sub-polyominoes. The type of polyomino shown above is a nice combination of these two properties.
Best Answer
On Torsten Sillke's site you can find results for all the pentacubes, including the Z-pentacube for which 6x6x25 and 6x10x10 solutions are known. They were found by Yoshiya Wolf Shindo in 1997.
The 6x10x10 solution has 4-fold rotational symmetry around one axis, and 2-fold around the other axes, so it consists of 8 copies of a single shape of approximately size 3x5x5 that fit together to form the block.