I realize that elements in the 3rd period and above have the unfilled d orbitals to expand their octet, but in the case of boron for example, they say boron can only have 6 valence electrons. doesnt the 2s2 orbital along with the p orbitals classify boron as an 8 valence electron atom as opposed to a 6.

Anyone who, using the Lewis/octet rule model, says that boron can have only six valence electrons is either wrong or incomplete. Electrically neutral boron (no formal charges) has three valence electrons; it can then covalently bond to three other things, each contributing another valence electron for a total of six. Thus boron is formally trivalent, with "six valence electrons" in compounds.

But, like any other first-row main-group element, boron prefers an octet. In neutral, covalent compounds that is done by some form of e lectron donation from another atom; but boron can also form ions of the form R₄B-, such as the common organic anion tetraphenylborate. In such ions, boron is tetravalent like carbon (we say that it is isoelectronic with carbon) but because it has one less proton than carbon, it carries a negative formal charge.