In C, do braces act as a stack frame?

No, braces do not act as a stack frame. In C, braces only denote a naming scope, but nothing gets destroyed nor is anything popped off the stack when control passes out of it.

As a programmer writing code, you can often think of it as if it is a stack frame. The identifiers declared within the braces are only accessible within the braces, so from a programmer's point of view, it is like they are pushed onto the stack as they are declared and then popped when the scope is exited. However, compilers don't have to generate code that pushes/pops anything on entry/exit (and generally, they don't).

Also note that local variables may not use any stack space at all: they could be held in CPU registers or in some other auxiliary storage location, or be optimized away entirely.

So, the d array, in theory, could consume memory for the entire function. However, the compiler may optimize it away, or share its memory with other local variables whose usage lifetimes do not overlap.

The time during which the variable is actually taking up memory is obviously compiler-dependent (and many compilers don't adjust the stack pointer when inner blocks are entered and exited within functions).

However, a closely related but possibly more interesting question is whether the program is allowed to access that inner object outside the inner scope (but within the containing function), ie:

void foo() {
   int c[100];
   int *p;

   {
       int d[200];
       p = d;
   }

   /* Can I access p[0] here? */

   return;
}

(In other words: is the compiler allowed to deallocate d, even if in practice most don't?).

The answer is that the compiler is allowed to deallocate d, and accessing p[0] where the comment indicates is undefined behaviour (the program is not allowed to access the inner object outside of the inner scope). The relevant part of the C standard is 6.2.4p5:

For such an object [one that has automatic storage duration] that does not have a variable length array type, its lifetime extends from entry into the block with which it is associated until execution of that block ends in any way. (Entering an enclosed block or calling a function suspends, but does not end, execution of the current block.) If the block is entered recursively, a new instance of the object is created each time. The initial value of the object is indeterminate. If an initialization is specified for the object, it is performed each time the declaration is reached in the execution of the block; otherwise, the value becomes indeterminate each time the declaration is reached.

Your question is not clear enough to be answered unambiguously.

On the one hand, compilers don't normally do any local memory allocation-deallocation for nested block scopes. The local memory is normally allocated only once at function entry and released at function exit.

On the other hand, when the lifetime of a local object ends, the memory occupied by that object can be reused for another local object later. For example, in this code

void foo()
{
  {
    int d[100];
  }
  {
    double e[20];
  }
}

both arrays will usually occupy the same memory area, meaning that the total amount of the local storage needed by function foo is whatever is necessary for the largest of two arrays, not for both of them at the same time.

Whether the latter qualifies as d continuing to occupy memory till the end of function in the context of your question is for you to decide.

It's implementation dependent. I wrote a short program to test what gcc 4.3.4 does, and it allocates all of the stack space at once at the start of the function. You can examine the assembly that gcc produces using the -S flag.