**The situation is quite the same when $V$ is not free over $R$.** You only need an extra parameter, namely the dimension $n_1$ of $(1, 0)V$ over $K$. The equivalence classes of Hermitian forms over $V$ are in one-to-one correspondance with the Smith normal forms of $n_1$-by-$n_2$ matrices $A$ over $K$ where $n_2 = \dim_K((0,1)V)$, i.e., the rank of $A$ is a complete invariant.

Let us first observe that $R$ is a $K$-algebra via the embedding $k \mapsto (k, k)$ of $K$ into $R$.
We denote by $\sigma$ the switch involution of $R$, i.e, the $K$-automorphism which swaps the coordinates of $R$. We will say that $H: V \times V \rightarrow R$ is an *Hermitian form* over $R$ if the identities $H(rv + v',w) = rH(v, w) + H(v', w), H(v, rw + w') = \sigma(r)H(v, w) + H(v, w')$ and $H(v, w) = \sigma(H(w,v))$ hold for every $r \in R$ and $v, v', w, w' \in V$. Two Hermitian forms $H$ and $H'$ over $V$ are said to be *equivalent* if there is an $R$-automorphism $\phi$ of $V$ such that $H'(v,w) = H(\phi(v), \phi(w))$ for every $v,w \in V$.

Let us set $e_1 = (1, 0), e_2 = (0,1) \in R$ and $V_i = e_iV$ for $i = 1, 2$. Then we have $V = V_1 \oplus V_2$ as an $R$-module. An Hermitian form $H$ over $R$ is fully determined by its restriction to $V_1 \times V_2$. Indeed, the restriction of $H$ to $V_i \times V_i$ is zero (write $H(e_i v, w)$ in two different ways) for $i = 1,2$ and we obtain the restriction to $V_2 \times V_1$ from the restriction to $V_1 \times V_2$ by swapping arguments at the source and the target. The restriction of $H$ to $V_1 \times V_2$ is any $K$-bilinear map taking values in the $K$-algebra $R$. Hence it identifies with a matrix in $M_{n_1, n_2}(K)$ where $n_i =\dim_K(V_i)$ for $i = 1, 2$. Thus $H$ can be written as an $n$-by-$n$ matrix over $K$ of the form $\text{Mat}(A) \Doteq \pmatrix{0 & A \\ A^t & 0}$ with $n = n_1 + n_2$ and $A \in M_{n_1, n_2}(K)$.

An $R$-automorphism of $V$ is given by an $n$-by-$n$ matrix of the form $\pmatrix{P & 0 \\ 0 & Q}$ with $P \in GL_{n_1}(K), Q \in GL_{n_2}(K)$. Two Hermitian forms $\text{Mat}(A)$ and $\text{Mat}(B)$ are equivalent if and only if we can find $P \in GL_{n_1}(K), Q \in GL_{n_2}(K)$ such that $P^tAQ = B$ and $Q^tA^tP = B^t$, the second condition being redundant. Now it should be evident that the rank of $A$ over $K$ is a complete invariant of equivalence.