Is Wigner D matrix also a spherical tensor? Can one think of Wigner D-matrix $\mathcal D_{mm''}^{(l)}$ 
as a spherical tensor even in case when $m''\neq0$ ?
In other words does relation like 
\begin{equation}
  \tag{1}
  \mathcal D^\dagger(R_1)\mathcal D^{(l)}_{mm''}(R_2)\mathcal D(R_1)
  = \sum_{m'} D_{mm'}^{(l)*}(R_1)D^{(l)}_{m'm''}(R_2)
\end{equation}
hold? And if not, can one argue that this makes physically sense?
In case when $m''=0$ relation above holds, since 
$\mathcal D_{m0}^{(l)}(\phi, \theta,\gamma)\sim Y_l^{m*}(\theta,\phi)$, and for spherical harmonics we have
\begin{equation}
  \tag{2}
  \mathcal D^\dagger(R)Y_l^{m}(\hat n)\mathcal D(R)
  = \sum_{m'} D_{mm'}^{(l)*}(R)Y_l^{m'}(\hat n).
\end{equation}
For a general case, I've tried using the product rule,
and I got close but not quite there:
\begin{equation}
  \tag{3}
  D_{mm''}^{(l)}(R_1R_2)
  =\sum_{m'} D_{mm'}^{(l)}(R_1) D_{m'm''}^{(l)}(R_2),
\end{equation}
but I'm not quite sure if I can interpret l.h.s here 
as $D_{mm''}^{(l)}(R_1R_2)
= \mathcal D^\dagger(R_1)D_{mm''}^{(l)}(R_2)\mathcal D(R_2)$.
But, even then it is not quite right since I'd need a conjugate on the r.h.s, i.e. $D_{m''m'}^{(l)*}(R_1)$ instead of $D_{m''m'}^{(l)}(R_1)$.
Than again, in case my interpretation of $D_{mm''}^{(l)}(R_1R_2)$ is ok,
this has to reduce to spherical harmonic case again when $m''=0$. 
So I'm left a bit confused... any help would be much appreciated. 
 A: I can't think of what it might mean to say that the $\mathfrak{D}^{(\ell)}$ matrices are spherical tensors.  But they are certainly related.  The best source I know of for this kind of thing is Thorne's 1980 review, in which he shows how various presentations relate to each other.  For example, he shows that the various common vector and tensor harmonics can be written by "coupling" the scalar spherical harmonics to basis vectors.  For example, using the usual Clebsch-Gordan coefficients, we can write
\begin{equation}
  Y^{\ell', \ell m}(\theta, \phi) = \sum_{m', m''} (1\ell'm''m' | \ell m) \boldsymbol{\xi}^{m''} Y^{\ell'm'},
\end{equation}
where
\begin{equation}
  \boldsymbol{\xi}^0 = \mathbf{e}_z
  \quad \text{and} \quad
  \boldsymbol{\xi}^{\pm 1} = \mp (\mathbf{e}_x \pm i \mathbf{e}_y) / \sqrt{2}.
\end{equation}
He also shows how these can be written in terms of symmetric trace-free tensors, which are another useful presentation that shows up especially in older work.  Alternatively, related vector and tensor harmonics can be written in terms of derivatives of the $Y^{\ell m}$.
A related idea is the spin-weighted spherical harmonic (SWSH).  These generalize the formula that you gave above for standard spherical harmonics, and are sometimes defined as
\begin{equation}
  {}_{s}Y_{\ell m}(\theta, \phi) = (-1)^s \sqrt{ \frac{2\ell+1} {4\pi}} \mathfrak{D}^{(\ell)}_{-s, m}(\phi, \theta, 0).
\end{equation}
Unlike "scalar" spherical harmonics, SWSHs don't transform among themselves under rotation, because there's an extra factor of $e^{i s \gamma}$ floating around (where $\gamma$ is that third Euler angle, which was set to $0$ in the equation above).  I actually wrote a paper on essentially this topic, in which I showed that SWSHs really need to be understood as functions on the rotation group [or more generally on $SU(2)$], instead of the sphere.  Then, and only then, do they correctly transform among themselves under rotations.  (Except, of course, when $s=0$ — in which case the extra factor floating around is just $1$ and the SWSH is equal to the usual spherical harmonic.)
However, as Thorne shows, standard SWSHs can also be "coupled" to basis vectors, and do end up resulting in spherical tensors again.  Basically, if your basis vectors are also poorly behaved, so that they transform with a factor of $e^{-i s \gamma}$, you can cancel out the misbehavior of spin-weighted spherical harmonics and get something that does actually transform correctly as a tensor.

Transformation of the $\mathfrak{D}^{(\ell)}$ matrices
After the discussion in the comments, it seems that the question really is: How do the Wigner-$\mathfrak{D}$ matrices transform under rotation?  To be specific, suppose we have some $R_1$, we choose some $\ell$, and now we can write down the actual numerical values of $\mathfrak{D}^{(\ell)}_{m, m'}(R_1)$.  But those numerical values depend on the basis that we chose for the vector space that is implicitly present in the definition of $R_1$.  So now we wonder what would happen to all those values if we rotated the vector basis by some other rotation — say $R_2$.
So first we have to figure out how rotations themselves — in particular $R_1$ — transform under rotations.  This isn't hard: rotations form a group, where the group operation is composition.  If you think of rotations in terms of the rotation matrix, composition is just matrix multiplication.(*)  So the rotated version of $R_1$ is just $R_1 R_2$.
Now knowing this, the answer is already contained in the question above.  The numerical values of the rotated $\mathfrak{D}^{(\ell)}_{m, m'}(R_1)$ matrix are just the values of the matrix of the rotated rotation:
\begin{equation}
  \tag{A}
  \mathfrak{D}^{(\ell)}_{m, m'}(R_1\, R_2)
  =
  \sum_\mu \mathfrak{D}^{(\ell)}_{m, \mu}(R_1) \mathfrak{D}^{(\ell)}_{\mu, m'}(R_2).
\end{equation}
This is the transformation law for the Wigner-$\mathfrak{D}$ matrices.
You can also extend this as needed for multiple rotations.  For example,
\begin{equation}
  \mathfrak{D}^{(\ell)}_{m, m'}(R_1\, R_2\, R_3)
  =
  \sum_{\mu,\mu'} \mathfrak{D}^{(\ell)}_{m, \mu}(R_1) \mathfrak{D}^{(\ell)}_{\mu, \mu'}(R_2) \mathfrak{D}^{(\ell)}_{\mu', m'}(R_3).
\end{equation}
So if, for some reason, you wanted to conjugate your original rotation, you would have
\begin{equation}
  \mathfrak{D}^{(\ell)}_{m, m'}(R_2\, R_1\, R_2^{-1})
  =
  \sum_{\mu,\mu'} \mathfrak{D}^{(\ell)}_{m, \mu}(R_2) \mathfrak{D}^{(\ell)}_{\mu, \mu'}(R_1) \mathfrak{D}^{(\ell)}_{\mu', m'}(R_2^{-1}).
\end{equation}
It's also worth noting that equation (A) tells us that the $\mathfrak{D}$ matrix for an inverse rotation is just the matrix-inverse of the original $\mathfrak{D}$ matrix:
\begin{equation}
  \mathfrak{D}^{(\ell)}_{m, m'}(R^{-1})
  =
  \left( \mathfrak{D}^{(\ell)}_{m, m'}(R) \right)^{-1}.
\end{equation}

(*) There's a little bit of subtlety about whether you mean your rotations to be active or passive, etc.  Depending on what exactly your conventions are, the correct result might be $R_2\, R_1$.  You need to work this out for yourself by actually rotating, say, the basis vectors and deciding what you personally mean by all these things.  If there's one thing I've learned in years of working on this stuff, it's that there is no universal agreement on literally anything when it comes to rotations.  For example, even the words "active" and "passive" can refer to totally different concepts, and different people might mean opposite things even when they're referring to the same concept.  So the most important rule is to check the basics for yourself.
A: I'm not sure about your notation but as I understand it I would start by writing
$$
D^\ell_{mm'}(R_2)=
\langle \ell m\vert R_2\vert \ell m'\rangle\, .
$$
Then 
\begin{align}
(R_1)^{-1}D^\ell_{mm'}(R_2)R_1&:=
\langle \ell m\vert R_1^{-1}R_2R_1\vert  \ell m'\rangle\, , \\
&=\sum_{\mu\mu'}
\langle \ell m\vert R_1^{-1} \vert \ell \mu\rangle 
\langle \ell \mu \vert R_2 \vert \ell \mu'\rangle
\langle \ell \mu'\vert R_1 \vert\ell m'\rangle\, , \\
&=\sum_{\mu\mu'}D^{\ell}_{m'\mu }(R^{-1}_1)D^{\ell}_{\mu \mu'}(R_2)
D^\ell_{\mu' m'}(R_1)
\end{align}
so it doesn't transform as a tensor because of the double sum.
