Can AES-CTR add values without decryption?
No, with AES in CTR mode, it is not


  possible to add/subtract(or XOR?) plaintext values on encrypted data


nor define operator $\boxplus$ such that $\operatorname{AES-CTR}(4)\boxplus\operatorname{AES-CTR}(6)=\operatorname{AES-CTR}(4+6)$ that works for unknown key and plaintexts. Informally: if that worked, it would be a weakness of the encryption scheme, which would be leaking information about plaintexts from the ciphertexts.

Notice that the above notation is misleading, as the result of AES-CTR depends on 3 parameters: Key, IV and plaintext, with IV normally random (making the argument above non-rigorous). But still, even if we assume fixed IV, what's asked can't be done: it is asking an operator $\boxplus$ such that $\forall s,\forall x,\forall y, (s\oplus x)\boxplus(s\oplus y)=s\oplus(x+y)$ where $\oplus$ is XOR, and $s$ is whatever the IV encrypts to under the key considered; which is impossible.

More generally, no secure mode of operation of any secure block cipher allow homomorphic encryption.