No. Assume that $*$ is definite,  that is, $\sum a_ia_i* = 0$ implies all $a_i = 0$ (which is obviously necessary). Necessary & sufficient for there to be a unital $*$-homomorphism to a $B(H)$ is that for all $a \in A$, there should exist finite $a_i \in A$ such that $aa*+\sum a_i a_i* $ is a (complex) scalar multiple of the identity. There are lots of examples of complex Baer $*$-algebras for which this fails (including ones for which the involution is definite, see below for the definition), e.g., the regular ring of a finite W*-factor. 

In general, the bounded subalgebra of a Baer $*$-algebra (the set of elements $a$ for which such $a_i$ exist; it is a $*$-subalgebra) admits a seminorm, which after its kernel is factored out, can be completed to a C*-algebra; hence sufficiency of the criterion—however, in general, we don't get that the image in a $B(H)$ is a Baer * subring.

In the finite-dimensional case, assuming the involution is definite, the algebra must be semisimple, and then the involution can be converted to a standard one.