The question has opposite answers, depending on if we consider adversaries computationally bounded, or not.
Actual adversaries are computationally bounded, that is have limited computational resources. Against these, any cipher that is secure against Chosen Plaintext Attack has the property that similarities between plaintexts are undetectable and unexploitable: the adversary gains no useful information (about the key or the plaintexts) from such similarities. With a properly chosen key, and a correct implementation (no side channel leakage, secure Random Number Generator for Initialization Vector, salt, …), that's the case for all modern ciphers, including the ciphers in the question:
- AES in any of the standard modes of operation except Electronic Code Book.
- RSA encryption per one of the PKCS#1 mode (RSAES-OAEP and the legacy RSAES-PKCS1-v1_5), or using an otherwise secure hybrid encryption. Textbook RSA $c_i\gets {t_i}^e\bmod n$ is not CPA-secure.
- The scrypt tool, which really uses AES in counter mode¹. For this tool, "properly chosen key" means that the entropy in the password, combined with the workfactor parameters used, must make brute force password search far exceed the computational capabilities of adversaries.
Against computationally unbounded adversaries, practical cryptographic schemes including those in the question become insecure no later than when what's known about plaintext exceeds the key entropy. In that case, learning that $k$ original plaintexts are mostly identical in a certain location reveals $k-1$ times the length of the common plaintext, and when that exceeds the key size it becomes theoretically possible to find the key, by enumeration of all keys, decrypting the ciphertexts, and keeping the (probably, single) key that makes all the tentatively decrypted plaintexts match in the stated location.
¹ According to this description of the format, the bulk of the ciphertext is data xor AES256-CTR key stream generated with nonce == 0. The later means that a multi-targets attack applies (assuming redundancy in plaintext, it is possible to test if a 256-bit AES key matches any of the ones used in multiple ciphertexts using a single block encryption), but that's unlikely to be a practical weakness.