Partial answer

First, let us define some things:

• For simplicity, let's look always for boards where the width is equals to or larger than the height. If not, you can just rotate everything 90° to get one that is.
• Unsolvable board (UB) - One that no matter what you rotates, it is impossible to have all the dominoes with the same orientation.
• Smallest unsolvable dimension (SUD) - The smallest number possible as the dimension of an UB.
• Minimal unsolvable board (MUB) - An UB that has one of its dimension sized the SUD and the other dimension as smallest as possible for it being an UB.
• Impossible board (IB) - A board that can't be filled with dominoes. Like one that has an odd number of tiles.

Lemma 1. An UB exists if, and only if, a MUB exists.

If there is an UB (or a lot of them), there is at least one that is also a MUB. Trivially, if there is a MUB, so there is some UB because every MUB is also an UB.

This also means that if no MUB exist, so no UB exist either. If no UB exist, there can't be a MUB also.

Lemma 2. A solvable board that have both dimensions even can be solved with either orientation.

Proof: Solve the board with any orientation. Split it into many $2 \times 2$ subblocks and rotate all of them. The result is a board solved to the other orientation.

Lemma 3. A solvable board that have an odd dimension must be solved with the dominoes oriented along the other dimension.

Proof: By being odd, you can't fill it with dominoes that are even-sized along this dimension.

Lemma 4. A MUB can't be splitted in two rectangular subboards along either dimension.

Let's suppose we have a MUB that is splittable in subboards like this:

╔═══...═══╗ ╔═...═╦═...═╗
║         ║ ║     ║     ║
...     ... ║     ║     ║
║         ║ ║     ║     ║
╠═══...═══╣ ...  ...  ...
║         ║ ║     ║     ║
...     ... ║     ║     ║
║         ║ ║     ║     ║
╚═══...═══╝ ╚═...═╩═...═╝

If one of the subboards is unsolvable, then the big board was not a MUB. If any subboard is an IB, so is the big board, so it is not a MUB either. So, both subboards must be solvable.

However, if both subboards are solvable, then if solved in the same orientation, the great board would also be solvable, hence not a MUB.

So, we have that the solvable subboards must be solvable in different orientations. Considering lemma 2, one of them must have $odd \times even$ and the other $even \times odd$ dimensions. There is no way to divide the big board to have this result because the length in which it is divided can't be odd and even at the same time.

The conclusion is that a MUB can't be divided in two rectangular subboards in any way. This lead us to the following lemma:

Lemma 5. A MUB necessarily features at least one domino spanning every pair of neighboring rows and every pair of neighbouring columns.

Because if it isn't, we will violate lemma 4.

Lemma 6. With one dimension sized 2, it is always solvable.

Proof: Let's start filling the top-left corner of the MUB with a SUD = 2. The two first possibilities are those:

╔═══╦═    ╔═╦═   
╠═══╣ ... ║ ║ ...
╚═══╩═    ╚═╩═   

By invoking lemma 4, those aren't MUBs. By invoking lemma 1, if there isn't a MUB, so there is not UB also. So, with dimension 2, it is always solvable.

Lemma 7. If there is a strip of dominoes where both sides have equally curved edges and it is not the entire board, then it is not a MUB either.

Let's consider an area like this:

    ═╦═══════╦══════  
     ╚═╗     ╚═╗  
    A  ╚═╗ B   ╚═╗ C    
...  ╔═══╝   ╔═══╝   ...
     ║       ║
    ═╩═══════╩══════

We see that the jagged line that divides $A$ from $B$ and $B$ from $C$ are equally curved. This means that we can delete $B$ and join $A$ and $C$. Thus $A+B+C$ is solvable if, and only if, $A+B$ is solvable.

Also, $A+B+C$ being an UB implies that $A+B$ is also an UB. So, $A+B+C$ is not a MUB.

Lemma 8. If there is a strip of dominoes, even if not contiguous, where both sides have equally curved edges and it is not the entire board, then it is not a MUB either.

Roughly the same as lemma 7, but with uncontiguous areas:

    ═╦═══════╦═══════════════  
     ╚═╗     ╚═╗  
       ╚═╗ B1  ╚═╗ C    
...      ╚═══════╩═╦═══════╗ ...
            A      ║  B2   ║
    ═══════════════╩═══════╩═

We might delete both the areas $B1$ and $B2$ and fit $A$ into $C$.

Lemma 9. With one dimension sized 3, it is always solvable.

Proof: Let's start filling the top-left corner of a MUB with one dimension sized 3. The two first possibilities are those:

╔═══╦═    ╔═╦═╦═   
╠═══╣ ... ║ ║ ║ ...
╠═══╣     ╠═╩═╣    
╚═══╩═    ╚═══╩═

Considering lemma 4, this can't be a MUB.

Let us try this other configuration:

╔═╦═══╦═══   
║ ╠═══╣   ...
╠═╩═╦═╩═╗   
╚═══╩═══╩═

By removing the three tiles around the top-left one and fitting it in the hole that would result, we get a smaller board, meaning again that it was not a MUB. This is also an application of lemma 7.

Hence, there is no MUB with a dimension sized 3, so $SUD > 3$.

Lemma 10. With one dimension sized 4, it is always solvable.

Filling the left side of a MUB:

╔═╦═    ╔═══╦═    ╔═╦═╦═    ╔═╦═══╦═    ╔═══╦═══    ╔═══╦═    ╔═══╦═══
║ ║     ╠═══╣     ║ ║ ║     ║ ╠═══╣     ╠═══╣       ╠═╦═╣     ╠═╦═╩═╗
╠═╣ ... ╠═══╣ ... ╠═╩═╣ ... ╠═╩═╦═╝ ... ╠═╦═╩═╗ ... ║ ║ ║ ... ║ ╠═══╣ ...
║ ║     ╠═══╣     ╠═══╣     ╠═══╣       ║ ╠═══╣     ╠═╩═╣     ╠═╩═╦═╝  
╚═╩═    ╚═══╩═    ╚═══╩═    ╚═══╩═══    ╚═╩═══╩═    ╚═══╩═    ╚═══╩═══
  A        B         C          D           E          F         G

Well, $A$, $B$, $C$ and $F$ violates lemma 4, so they aren't MUBs. $D$, $E$ and $G$ violates lemma 7, so they aren't either.

Since there isn't a MUB with the smallest dimension as 4 and from lemmas 8 and 9, not with smaller sizes, then $SUD > 4$.