Let $M$ be a $R$-Bimodule that happens to be projective, is its associated left $R \otimes R^{op}$-module projective too? Let $k$ be a commutative ring, a $R$-Bimodule $M$ over a $k$-algebra $R$ is a $k$-module with two actions of $R$ on $M$, on the left and on the right, the classical example of this being $R$ itself with left and right multiplications as the actions. A $R$-Bimodule $M$ can also be considered a left $R \otimes R^{op}$-module.
Now what I want to know:
1 - Let this $R$-Bimodule $M = R$, let $R$ (considered as a left $k$-module) be projective (actually, let $k$ be a field, so $R$ is a vector space), is the corresponding $R \otimes R^{op}$-module a projective module too?
2 - 1 - Let this $R$-Bimodule $M = R$, let $R$ (considered as a left $k$-module) be free (actually, let $k$ be a field, so $R$ is a vector space), is the corresponding $R \otimes R^{op}$-module a free module too?
 A: No. A ring $R$ for which $R$ is projective as a left $R\otimes_{\mathbb Z}R^{op}$-module is sometimes called separable over ${\mathbb Z}$.
This is equivalent to the splitting, as a surjection of $R\otimes_{\mathbb Z}R^{op}$-modules, of the multiplication map $R\otimes_{\mathbb Z}R\to R$. So applying $-\otimes_R M$, the $R$-action map $R\otimes_{\mathbb Z}M\to M$ has to split as a surjection of $R$-modules for every left $R$-module $M$. In particular, if $M$ is ${\mathbb Z}$-free, so that $R\otimes_{\mathbb Z}M$ is free as an $R$-module, $M$ must be projective as an $R$-module.
But it's not true that every ${\mathbb Z}$-free $R$-module is projective if (as in Steven's answer) $R=\mathbb{Z}[X]$ or $R=\mathbb{Z}G$ for a non-trivial group $G$, or a host of other examples. 
A: I'm not sure what hypothesis you intend, but I don't think there's any reasonable interpretation under which it implies your conclusion.
Let $R=M={\mathbb Z}[X]$, acting on itself by multiplication from both the left and the right.  Then $R\otimes R^{op}\approx {\mathbb Z}[X,Y]$ is a domain, and $(X-Y)$ annihilates $M$, whence $M$ cannot be free.  
