No, knowing the SHA-256 hash of an arbitrary message, and a key, it is not possible to compute the HMAC-SHA-256 for that message and key; that's because in HMAC, something dependent on the key is put in front of the message before hashing the message.
Addition: if what's asked was possible, then a break of SHA-256 allowing to find collisions would be a death blow for HMAC-SHA-256 (the HMAC for messages colliding per SHA-256 would be identical for any key). By contrast, we have excellent hope that the HMAC construction would provide enough protection to keep HMAC-SHA-256 practically safe. There's the precedent of HMAC-MD5, which remains unbroken even though MD5's collision resistance is trounced; and the theoretical security argument made by Mihir Bellare in New Proofs for NMAC and HMAC: Security without Collision-Resistance.
Independently, on the same line: knowing the SHA-256 hash of an arbitrary key of more than 64 bytes, and a message, it is possible to compute the HMAC-SHA-256 for that message and key; that's because by definition of HMAC, for keys larger than the message block size (64 bytes for SHA-256), the first operation made with the key is to replace it by its hash.
Note: in this answer, arbitrary excludes guessing a low-entropy value from its hash.