There is a famous result about perfect secrecy, usually called Shannon's theorem:
If an encryption scheme has perfect secrecy, then $|\mathcal{K}| \ge |\mathcal{M}|$, where $\mathcal{K}$ is the set of possible keys and $\mathcal{M}$ is the set of possible plaintexts.
One-time pad achieves equality $|\mathcal{K}| = |\mathcal{M}|$, which is the fewest number of keys possible according to Shannon's theorem.
If you seek a scheme that is "no less practical" than OTP, then I suppose that you too require $|\mathcal{K}| = |\mathcal{M}|$.
OTP also has a deterministic encryption algorithm, so any scheme that is "no less practical" should also have deterministic encryption.
(Also, randomized encryption would needlessly increase the ciphertext size.)
With a deterministic encryption scheme, we can imagine writing out the entire truth table of $\textsf{Enc}(K,M)$, with rows indexed by $K \in \mathcal K$ and columns indexed by $M = \mathcal M$.
As $|\mathcal M|=|\mathcal K|$, the table is a square.
For decryption to be possible, every row must have distinct entries -- this corresponds to the property that $\textsf{Enc}(K,\cdot)$ is a bijection for every $K$.
Every column must also have distinct entries: if the column corresponding to $M$ does not have distinct entries, then it is missing some $C$, meaning that no key encrypts $M$ to $C$.
But then observing the ciphertext $C$ means the plaintext could not have been $M$, and this violates perfect secrecy.
(We know $C$ appears somewhere in the table as encryption of some plaintext, otherwise the condition on rows doesn't hold.)
The truth table of $\textsf{Enc}$ is therefore a Latin square, making it the truth table of a quasigroup operation.
Let's call the quasigroup $(\mathcal G, \boxplus)$.
Then $\mathcal M$ and $\mathcal K$ are both isomorphic to $\mathcal G$ and the encryption algorithm is $\textsf{Enc}(K,M) = K \boxplus M$.
So the question of alternative perfectly secret encryption schemes reduces to the question of identifying a suitable quasigroup.
