I am not a cipher guru, but an obvious question comes to mind: would you be allowed to use One Time Pad encryption? Then you can just include a large block of truly random bits in your decoding system, and use the random data to transform your decimal digits in a reversible way.
If this would be acceptable, we just need to figure out how the decoder knows where in the block of randomness to look to get the key to decode any particular message. If you can send a plaintext timestamp with the ciphertext, then it's easy: convert the timestamp into a number, say the number of seconds since an epoch date, modulus that number by the length of the randomness block, and you have an offset within the block.
With a large enough block of randomness, this should be uncrackable. You could have the random bits be themselves encrypted with strong encryption, such that the user must type in a long password to unlock the decoder; in this way, even if the decryption software was captured, it would still not be easy to break the system.
If you have any interest in this and would like me to expand further, let me know. I don't want to spend a lot of time on an answer that doesn't meet your needs at all.
EDIT: Okay, with the tiny shred of encouragement ("you might be on to something") I'm expanding my answer.
The idea is that you get a block of randomness. One easy way to do this is to just pull data out of the Linux /dev/random
device. Now, I'm going to assume that we have some way to find an index into this block of randomness for each message.
Index into the block of randomness and pull out ten bytes of data. Each byte is a number from 0 to 255. Add each of these numbers to the respective digit from the plaintext, modulo by 10, and you have the digits of the ciphertext. You can easily reverse this as long as you have the block of random data and the index: you get the random bits and subtract them from the cipher digits, modulo 10.
You can think of this as arranging the digits from 0 to 9 in a ring. Adding is counting clockwise around the ring, and subtracting is counting counter-clockwise. You can add or subtract any number and it will work. (My original version of this answer suggested using only 3 bits per digit. Not enough, as pointed out below by @Baffe Boyois. Thank you for this correction.)
If the plain text digit is 6, and the random number is 117, then: 6 + 117 == 123, modulo 10 == 3. 3 - 117 == -114, modulo 10 == 6.
As I said, the problem of finding the index is easy if you can use external plaintext information such as a timestamp. Even if your opponent knows you are using the timestamp to help decode messages, it does no good without the block of randomness.
The problem of finding the index is also easy if the message is always delivered; you can have an agreed-upon system of generating a series of indices, and say "This is the fourth message I have received, so I use the fourth index in the series." As a trivial example, if this is the fourth message received, you could agree to use an index value of 16 (4 for fourth message, times 4 bytes per one-time pad). But you could also use numbers from an approved pseudorandom number generator, initialized with an agreed constant value as a seed, and then you would get a somewhat unpredictable series of indexes within the block of randomness.
Depending on your needs, you could have a truly large chunk of random data (hundreds of megabytes or even more). If you use 10 bytes as a one-time pad, and you never use overlapping pads or reuse pads, then 1 megabyte of random data would yield over 100,000 one-time pads.