If a matrix $A$, not necessarily symmetric, has real, nonnegative eigenvalues, is it positive semidefinite?

No, consider the following matrix:

$$A=\pmatrix{1&-10\\0&2}$$

It has positive eigenvalues, but is certainly not positive semidefinite.

The answer to your question is NO. (And it is worse in the complex case: if a matrix $A$ is not Hermitian, then it is impossible that $x^TAx\geq 0$ for all $x$.)

In the setting of complex vector spaces, if $x^TAx\geq 0$ for all complex vectors $x$ (your definition of semi-definite), which in particular implies that $x^TAx\in{\bf R}$, the following theorem in Axler's Linear Algebra Done Right shows that $A$ must be Hermitian (if $A$ is real, then $A$ must be symmetric):

In the setting of real vector space, one has the following simple counterexample.

For any $(x,y)\in{\bf R}^2$, $$ (x,y)\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}=xy. $$

Given that, for a real matrix $A$ we have $$ x^T A x=x^T A^{(S)} x $$ where $A^{(S)}$ is the symmetric part of $A$, then the character of $A$ with respect to positive definiteness, or semi-definiteness, is related only to the eigenvalues of its symmetric part.

As the examples in other answers show, a matrix could have positive eigenvalues, but its symmetric part could have a negative eigenvalue, so eigenvalues of a matrix could not be related to positive (semi)definiteness.