Naive RSA is not secure. Even with some kinds of padding, it's not
secure. Allow me to quote from Bodo Muller's recent posting on sci.crypt:
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A padding scheme that is deterministic cannot be
secure even against passive adversaries (assuming that you want at
least semantic security): If the adversary can guess what the
plaintext might be, then he can easily verify if this guess is true
just by encrypting the message himself and comparing the result with a
given ciphertext.

So there are probabilistic padding schemes such as the one from
PKCS v#1.5 that add a large number of random bits and some fixed bits.
But even these have vulnerabilities in the presence of active
adversaries: For example, Bleichenbacher's attack on RSA with PKCS
#v1.5 can decrypt arbitrary ciphertexts; it requires just an oracle
that tries to decrypt queries and merely reveals if this decryption
attempt succeeded or not (i.e., whether the result of the exponentiation
with the RSA private key did contain the expected fixed bits specified
by the padding scheme).

Thus more sophisticated padding schemes are called for.

OAEP, "Optimal Asymmetric Encryption Padding", was the first one with
a security proof.  The proof later turned out not to prove what it was
expected to prove, but RSA-OAEP was later proven secure anyway.  The
remaining problem is that the RSA-OAEP security proof isn't really all
that meaningful: It says essentially "if you can break RSA-OAEP, then
with some effort you can compute e-th roots modulo n" (presumably the
latter is hard, which would appear to imply that RSA-OAEP must be
secure).  But at typical key sizes, the specific effort needed here is
more than would be required to go straight ahead and factor n and
break RSA that way instead.

But OAEP led to further padding schemes: by now we have OAEP+, SAEP,
and SAEP+.
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  Meaning, it need not interoperate with other encryption applications?

  If that's a deterministic transformation of the ciphertext, it
doesn't affect security.

--Mike Amling