Why are rotational matrices not commutative? Is there any intuition why rotational matrices are not commutative? I assume the final rotation is the combination of all rotations. Then how does it matter in which order the rotations are applied?
Here is a picture of a die:



Now let's spin it $90^\circ$ clockwise. The die now shows



After that, if we flip the left face up, the die lands at





Now, let's do it the other way around: We start with the die in the same position:



Flip the left face up:



and then $90^\circ$ clockwise



If we do it one way, we end up with $3$ on the top and $5, 6$ facing us, while if we do it the other way we end up with $2$ on the top and $1, 3$ facing us. This demonstrates that the two rotations do not commute.



Since so many in the comments have come to the conclusion that this is not a complete answer, here are a few more thoughts:


Note what happens to the top number of the die: In the first case we change what number is on the left face, then flip the new left face to the top. In the second case we first flip the old left face to the top, and then change what is on the left face. This makes two different numbers face up.
As leftaroundabout said in a comment to the question itself, rotations not commuting is not really anything noteworthy. The fact that they do commute in two dimensions is notable, but asking why they do not commute in general is not very fruitful apart from a concrete demonstration.
I'm discussing here non-commutativity of rotations made about current axes..

Rotations in 3d are non commutative because  rotation changes direction of every potential other axis except itself (unlike in 2d where it is nothing to change because it is only one "axis" of rotation - it can be reduced in 3D to rotation about Z axis ). 

Denote succession of rotations about Z-axis as $Z_1,  Z_2,... Z_n$ (it's convenient to imagine that there is multitude of them by a small angular increment let's say $1^\circ$ or $2^\circ$ ) and one rotation about different axis $X$ ( it can be also not very big).  When we have only rotations $..Z_j...$ we can change order of rotations in any way - the result will be the same.

But if  we  insert  rotation $\mathbf{X}$ somewhere into middle of rotations succession   $Z_1,{Z_2}...{Z_n}$ the final result $Z_1,{z_2}...{Z_j}{\mathbf{X}}  {Z_k}...{Z_n}$ now depends on which place we inserted it because inserting it means that we are changing current axis of rotations for all successive rotations Z - the current Z axis, after this operation, is changed in the global frame.

It could be seen the other effect of inserting X rotation into different places of succession of $ Z_i $ if we would track the end of X versor (vector $i$) : 


If this X rotation is made at the beginning the plane XY constant in
current frame for Z rotations is changed and the endpoint of X axis
is distancing itself from original XY plane with every successive
incremental rotation about changed now Z-axis.
If it would be made at the end, the distance from original XY plane
still would be equal $0$ !

As the versors $i,j,k$  after whole linear operation represents the resulting matrix we clearly see that these two operations are quite different.
Graphically one can imagine this track of the endpoint of $X$ versor on the surface of the unit sphere.




The simplest alternative presentation of non-commutativity in 3D (approach here is more algebraic than geometric as previous one) can be made using axis unit vector $v_1$ of   rotation, say, $R_1$. 

($R_2$, we assume, has different axis with vector $v_2 \ne v_1$ and moreover we assume $R_2(v_1) \ne -v_1$ which together with $R_2(v_1) \ne  v_1$ states that transformed by $R_2$ vector $v_1$ is not lying in the axis of $R_1$).  

Apply it to $R_2R_1(v_1)=R_2(v_1) = v_{21}$. (Obviously $v_{21} \ne v_1$).  

In reverse order $R_1R_2(v_1)=R_1(v_{21} )$.
But  $v_{21} $ is not lying in the axis of $R_1$ !
Hence the result must be different than $v_{21}$ and $R_1R_2 \ne R_2R_1$.    

They commute only if they share common axis or in the case of different axes they preserve each other axes with result vector  changing sign i.e. $R_2(v_1)=-v_1$ and $R_1(v_2)=-v_2$ what is possible when $v_1$ is orthogonal to $v_2$ and rotations are by $\pi$ angle.